Let’s calculate the final temperature step by step.<\/p>\n\n\n\n
Step 1: Define the Given Data<\/strong><\/h3>\n\n\n\n\n- Mass of ice cubes = 2\u00d720.02 \\times 20.0 g = 40.0 g<\/strong><\/li>\n\n\n\n
- Initial temperature of ice = -17.00\u00b0C<\/strong><\/li>\n\n\n\n
- Mass of water = 470.1 g<\/strong><\/li>\n\n\n\n
- Initial temperature of water = 25.00\u00b0C<\/strong><\/li>\n\n\n\n
- Heat of fusion of ice = 6.008 kJ\/mol<\/strong><\/li>\n\n\n\n
- Molar heat capacity of ice = 37.7 J\/mol\u00b7K<\/strong><\/li>\n\n\n\n
- Molar heat capacity of water = 75.3 J\/mol\u00b7K<\/strong><\/li>\n\n\n\n
- Molar mass of water (H\u2082O) = 18.015 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
Step 2: Convert Mass to Moles<\/strong><\/h3>\n\n\n\nMoles of ice=40.0 g18.015 g\/mol=2.22 mol\\text{Moles of ice} = \\frac{40.0 \\text{ g}}{18.015 \\text{ g\/mol}} = 2.22 \\text{ mol}<\/p>\n\n\n\n
Step 3: Calculate Energy to Warm Ice to 0\u00b0C<\/strong><\/h3>\n\n\n\nq1=moles of ice\u00d7Cice\u00d7\u0394Tq_1 = \\text{moles of ice} \\times C_{\\text{ice}} \\times \\Delta T q1=(2.22 mol)\u00d7(37.7 J\/mol\\cdotpK)\u00d7(0\u2212(\u221217.00))q_1 = (2.22 \\text{ mol}) \\times (37.7 \\text{ J\/mol\u00b7K}) \\times (0 – (-17.00)) q1=(2.22)\u00d7(37.7)\u00d7(17.00)q_1 = (2.22) \\times (37.7) \\times (17.00) q1=1411.6 J=1.41 kJq_1 = 1411.6 \\text{ J} = 1.41 \\text{ kJ}<\/p>\n\n\n\n
Step 4: Energy to Melt Ice<\/strong><\/h3>\n\n\n\nq2=moles of ice\u00d7\u0394Hfusq_2 = \\text{moles of ice} \\times \\Delta H_{\\text{fus}} q2=(2.22 mol)\u00d7(6.008 kJ\/mol)q_2 = (2.22 \\text{ mol}) \\times (6.008 \\text{ kJ\/mol}) q2=13.34 kJq_2 = 13.34 \\text{ kJ}<\/p>\n\n\n\n
Step 5: Energy Released by Water Cooling to TfT_f<\/strong><\/h3>\n\n\n\nq3=moles of water\u00d7Cwater\u00d7\u0394Tq_3 = \\text{moles of water} \\times C_{\\text{water}} \\times \\Delta T<\/p>\n\n\n\n
Moles of water: 470.1 g18.015 g\/mol=26.10 mol\\frac{470.1 \\text{ g}}{18.015 \\text{ g\/mol}} = 26.10 \\text{ mol}<\/p>\n\n\n\n
Total heat available from water cooling: q3=(26.10)\u00d7(75.3)\u00d7(25.00\u2212Tf)q_3 = (26.10) \\times (75.3) \\times (25.00 – T_f) q3=1965.3(25.00\u2212Tf) Jq_3 = 1965.3 (25.00 – T_f) \\text{ J} q3=1.965(25.00\u2212Tf) kJq_3 = 1.965 (25.00 – T_f) \\text{ kJ}<\/p>\n\n\n\n
Step 6: Solve for TfT_f<\/strong><\/h3>\n\n\n\nq1+q2=q3q_1 + q_2 = q_3 (1.41+13.34)=1.965(25.00\u2212Tf)(1.41 + 13.34) = 1.965 (25.00 – T_f) 14.75=1.965(25.00\u2212Tf)14.75 = 1.965 (25.00 – T_f) Tf=25.00\u221214.751.965T_f = 25.00 – \\frac{14.75}{1.965} Tf=25.00\u22127.51T_f = 25.00 – 7.51 Tf=17.49\u2218CT_f = 17.49^\\circ C<\/p>\n\n\n\n
Final Answer:<\/strong><\/h3>\n\n\n\n17.49\u2218C\\mathbf{17.49^\\circ C}<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nThe problem involves two ice cubes at -17.00\u00b0C placed into a container of liquid water at 25.00\u00b0C. To find the final temperature after all the ice melts, we need to consider the energy transfer.<\/p>\n\n\n\n
First, the ice warms up from -17.00\u00b0C to 0\u00b0C. Using the molar heat capacity of ice (37.7 J\/mol\u00b7K) and the moles of ice (2.22 mol), the energy required for this is 1.41 kJ<\/strong>.<\/p>\n\n\n\nNext, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\nThe heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n
By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\nSince no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n
This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\nNow, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n
Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-1012.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 2: Convert Mass to Moles<\/strong><\/h3>\n\n\n\nMoles of ice=40.0 g18.015 g\/mol=2.22 mol\\text{Moles of ice} = \\frac{40.0 \\text{ g}}{18.015 \\text{ g\/mol}} = 2.22 \\text{ mol}<\/p>\n\n\n\n
Step 3: Calculate Energy to Warm Ice to 0\u00b0C<\/strong><\/h3>\n\n\n\nq1=moles of ice\u00d7Cice\u00d7\u0394Tq_1 = \\text{moles of ice} \\times C_{\\text{ice}} \\times \\Delta T q1=(2.22 mol)\u00d7(37.7 J\/mol\\cdotpK)\u00d7(0\u2212(\u221217.00))q_1 = (2.22 \\text{ mol}) \\times (37.7 \\text{ J\/mol\u00b7K}) \\times (0 – (-17.00)) q1=(2.22)\u00d7(37.7)\u00d7(17.00)q_1 = (2.22) \\times (37.7) \\times (17.00) q1=1411.6 J=1.41 kJq_1 = 1411.6 \\text{ J} = 1.41 \\text{ kJ}<\/p>\n\n\n\n
Step 4: Energy to Melt Ice<\/strong><\/h3>\n\n\n\nq2=moles of ice\u00d7\u0394Hfusq_2 = \\text{moles of ice} \\times \\Delta H_{\\text{fus}} q2=(2.22 mol)\u00d7(6.008 kJ\/mol)q_2 = (2.22 \\text{ mol}) \\times (6.008 \\text{ kJ\/mol}) q2=13.34 kJq_2 = 13.34 \\text{ kJ}<\/p>\n\n\n\n
Step 5: Energy Released by Water Cooling to TfT_f<\/strong><\/h3>\n\n\n\nq3=moles of water\u00d7Cwater\u00d7\u0394Tq_3 = \\text{moles of water} \\times C_{\\text{water}} \\times \\Delta T<\/p>\n\n\n\n
Moles of water: 470.1 g18.015 g\/mol=26.10 mol\\frac{470.1 \\text{ g}}{18.015 \\text{ g\/mol}} = 26.10 \\text{ mol}<\/p>\n\n\n\n
Total heat available from water cooling: q3=(26.10)\u00d7(75.3)\u00d7(25.00\u2212Tf)q_3 = (26.10) \\times (75.3) \\times (25.00 – T_f) q3=1965.3(25.00\u2212Tf) Jq_3 = 1965.3 (25.00 – T_f) \\text{ J} q3=1.965(25.00\u2212Tf) kJq_3 = 1.965 (25.00 – T_f) \\text{ kJ}<\/p>\n\n\n\n
Step 6: Solve for TfT_f<\/strong><\/h3>\n\n\n\nq1+q2=q3q_1 + q_2 = q_3 (1.41+13.34)=1.965(25.00\u2212Tf)(1.41 + 13.34) = 1.965 (25.00 – T_f) 14.75=1.965(25.00\u2212Tf)14.75 = 1.965 (25.00 – T_f) Tf=25.00\u221214.751.965T_f = 25.00 – \\frac{14.75}{1.965} Tf=25.00\u22127.51T_f = 25.00 – 7.51 Tf=17.49\u2218CT_f = 17.49^\\circ C<\/p>\n\n\n\n
Final Answer:<\/strong><\/h3>\n\n\n\n17.49\u2218C\\mathbf{17.49^\\circ C}<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nThe problem involves two ice cubes at -17.00\u00b0C placed into a container of liquid water at 25.00\u00b0C. To find the final temperature after all the ice melts, we need to consider the energy transfer.<\/p>\n\n\n\n
First, the ice warms up from -17.00\u00b0C to 0\u00b0C. Using the molar heat capacity of ice (37.7 J\/mol\u00b7K) and the moles of ice (2.22 mol), the energy required for this is 1.41 kJ<\/strong>.<\/p>\n\n\n\nNext, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\nThe heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n
By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\nSince no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n
This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\nNow, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n
Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
q1=moles of ice\u00d7Cice\u00d7\u0394Tq_1 = \\text{moles of ice} \\times C_{\\text{ice}} \\times \\Delta T q1=(2.22 mol)\u00d7(37.7 J\/mol\\cdotpK)\u00d7(0\u2212(\u221217.00))q_1 = (2.22 \\text{ mol}) \\times (37.7 \\text{ J\/mol\u00b7K}) \\times (0 – (-17.00)) q1=(2.22)\u00d7(37.7)\u00d7(17.00)q_1 = (2.22) \\times (37.7) \\times (17.00) q1=1411.6 J=1.41 kJq_1 = 1411.6 \\text{ J} = 1.41 \\text{ kJ}<\/p>\n\n\n\n
Step 4: Energy to Melt Ice<\/strong><\/h3>\n\n\n\nq2=moles of ice\u00d7\u0394Hfusq_2 = \\text{moles of ice} \\times \\Delta H_{\\text{fus}} q2=(2.22 mol)\u00d7(6.008 kJ\/mol)q_2 = (2.22 \\text{ mol}) \\times (6.008 \\text{ kJ\/mol}) q2=13.34 kJq_2 = 13.34 \\text{ kJ}<\/p>\n\n\n\n
Step 5: Energy Released by Water Cooling to TfT_f<\/strong><\/h3>\n\n\n\nq3=moles of water\u00d7Cwater\u00d7\u0394Tq_3 = \\text{moles of water} \\times C_{\\text{water}} \\times \\Delta T<\/p>\n\n\n\n
Moles of water: 470.1 g18.015 g\/mol=26.10 mol\\frac{470.1 \\text{ g}}{18.015 \\text{ g\/mol}} = 26.10 \\text{ mol}<\/p>\n\n\n\n
Total heat available from water cooling: q3=(26.10)\u00d7(75.3)\u00d7(25.00\u2212Tf)q_3 = (26.10) \\times (75.3) \\times (25.00 – T_f) q3=1965.3(25.00\u2212Tf) Jq_3 = 1965.3 (25.00 – T_f) \\text{ J} q3=1.965(25.00\u2212Tf) kJq_3 = 1.965 (25.00 – T_f) \\text{ kJ}<\/p>\n\n\n\n
Step 6: Solve for TfT_f<\/strong><\/h3>\n\n\n\nq1+q2=q3q_1 + q_2 = q_3 (1.41+13.34)=1.965(25.00\u2212Tf)(1.41 + 13.34) = 1.965 (25.00 – T_f) 14.75=1.965(25.00\u2212Tf)14.75 = 1.965 (25.00 – T_f) Tf=25.00\u221214.751.965T_f = 25.00 – \\frac{14.75}{1.965} Tf=25.00\u22127.51T_f = 25.00 – 7.51 Tf=17.49\u2218CT_f = 17.49^\\circ C<\/p>\n\n\n\n
Final Answer:<\/strong><\/h3>\n\n\n\n17.49\u2218C\\mathbf{17.49^\\circ C}<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nThe problem involves two ice cubes at -17.00\u00b0C placed into a container of liquid water at 25.00\u00b0C. To find the final temperature after all the ice melts, we need to consider the energy transfer.<\/p>\n\n\n\n
First, the ice warms up from -17.00\u00b0C to 0\u00b0C. Using the molar heat capacity of ice (37.7 J\/mol\u00b7K) and the moles of ice (2.22 mol), the energy required for this is 1.41 kJ<\/strong>.<\/p>\n\n\n\nNext, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\nThe heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n
By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\nSince no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n
This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\nNow, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n
Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
q3=moles of water\u00d7Cwater\u00d7\u0394Tq_3 = \\text{moles of water} \\times C_{\\text{water}} \\times \\Delta T<\/p>\n\n\n\n
Moles of water: 470.1 g18.015 g\/mol=26.10 mol\\frac{470.1 \\text{ g}}{18.015 \\text{ g\/mol}} = 26.10 \\text{ mol}<\/p>\n\n\n\n
Total heat available from water cooling: q3=(26.10)\u00d7(75.3)\u00d7(25.00\u2212Tf)q_3 = (26.10) \\times (75.3) \\times (25.00 – T_f) q3=1965.3(25.00\u2212Tf) Jq_3 = 1965.3 (25.00 – T_f) \\text{ J} q3=1.965(25.00\u2212Tf) kJq_3 = 1.965 (25.00 – T_f) \\text{ kJ}<\/p>\n\n\n\n
Step 6: Solve for TfT_f<\/strong><\/h3>\n\n\n\nq1+q2=q3q_1 + q_2 = q_3 (1.41+13.34)=1.965(25.00\u2212Tf)(1.41 + 13.34) = 1.965 (25.00 – T_f) 14.75=1.965(25.00\u2212Tf)14.75 = 1.965 (25.00 – T_f) Tf=25.00\u221214.751.965T_f = 25.00 – \\frac{14.75}{1.965} Tf=25.00\u22127.51T_f = 25.00 – 7.51 Tf=17.49\u2218CT_f = 17.49^\\circ C<\/p>\n\n\n\n
Final Answer:<\/strong><\/h3>\n\n\n\n17.49\u2218C\\mathbf{17.49^\\circ C}<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nThe problem involves two ice cubes at -17.00\u00b0C placed into a container of liquid water at 25.00\u00b0C. To find the final temperature after all the ice melts, we need to consider the energy transfer.<\/p>\n\n\n\n
First, the ice warms up from -17.00\u00b0C to 0\u00b0C. Using the molar heat capacity of ice (37.7 J\/mol\u00b7K) and the moles of ice (2.22 mol), the energy required for this is 1.41 kJ<\/strong>.<\/p>\n\n\n\nNext, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\nThe heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n
By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\nSince no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n
This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\nNow, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n
Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
17.49\u2218C\\mathbf{17.49^\\circ C}<\/p>\n\n\n\n
\n\n\n\n
Explanation <\/strong><\/h3>\n\n\n\nThe problem involves two ice cubes at -17.00\u00b0C placed into a container of liquid water at 25.00\u00b0C. To find the final temperature after all the ice melts, we need to consider the energy transfer.<\/p>\n\n\n\n
First, the ice warms up from -17.00\u00b0C to 0\u00b0C. Using the molar heat capacity of ice (37.7 J\/mol\u00b7K) and the moles of ice (2.22 mol), the energy required for this is 1.41 kJ<\/strong>.<\/p>\n\n\n\nNext, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\nThe heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n
By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\nSince no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n
This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\nNow, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n
Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Next, the ice melts at 0\u00b0C, requiring additional energy. The heat of fusion of ice is 6.008 kJ\/mol<\/strong>, and for 2.22 mol of ice, the energy needed to melt is 13.34 kJ<\/strong>.<\/p>\n\n\n\n The heat needed to warm the ice and melt it must come from the surrounding water, which cools down from 25.00\u00b0C to the final temperature TfT_f. The energy released by cooling the water is calculated using the molar heat capacity of water (75.3 J\/mol\u00b7K). The water cools until it has transferred enough heat to balance the total energy required by the ice.<\/p>\n\n\n\n By solving the heat balance equation, we find that the final equilibrium temperature is 17.49\u00b0C<\/strong>. This means the ice completely melts, and the entire system reaches thermal equilibrium at this temperature.<\/p>\n\n\n\n Since no heat is lost to the surroundings, the principle of conservation of energy holds: the heat gained by ice equals the heat lost by water.<\/p>\n\n\n\n This calculation confirms that after the ice melts, the final temperature of the system is 17.49\u00b0C<\/strong>.<\/p>\n\n\n\n Now, I’ll generate an image illustrating the process of heat transfer between ice and water.<\/p>\n\n\n\n Here is an illustration showing the heat transfer process as the ice melts in the water, leading to a final equilibrium temperature of 17.49\u00b0C. Let me know if you need any modifications!<\/p>\n\n\n\n Two 20.0 g ice cubes at -17.00C are placed into 470.1 g of water at 25.00C. The heat of fusion of ice is 6.008 kJ\/mol, the molar heat capacity of water is 75.3 J mol-1K-1 and the molar heat capacity of ice is 37.7 J mol-1K-1. Assuming no energy is transferred to or from the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201644","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201644"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201644\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGenerated Image<\/strong><\/h3>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"