{"id":194040,"date":"2025-02-22T07:25:08","date_gmt":"2025-02-22T07:25:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194040"},"modified":"2025-02-22T07:25:11","modified_gmt":"2025-02-22T07:25:11","slug":"17-4-for-aluminum-the-heat-capacity-at-constant-volume-cv-at-30-k-is-0-81-j-mol-k-and-the-debye-temperature-is-375-k","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/22\/17-4-for-aluminum-the-heat-capacity-at-constant-volume-cv-at-30-k-is-0-81-j-mol-k-and-the-debye-temperature-is-375-k\/","title":{"rendered":"17.4 For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J\/mol-K and the Debye temperature is 375 K"},"content":{"rendered":"\n

17.4 For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J\/mol-K and the Debye temperature is 375 K. Estimate the specific heat (a) at 50 K and (b) at 425 K. Solution (a) For aluminum, Cv at 50 K may be approximated by Equation 17.2, since this temperature is significantly below the Debye temperature (375 K). The value of Cv at 30 K is given, and thus, we may compute the constant A as A = Cv T 3 = 0.81 J\/mol-K (30 K) 3 = 3.00 \u00b4 10-5 J\/mol-K 4 Therefore, at 50 K Cv = AT 3 = (3.00 \u00b4 10-5 J\/mol-K 4 )(50 K) 3 = 3.75 J\/mol-K and cv = (3.75 J\/mol-K)(1 mol\/26.98 g)(1000 g\/kg) = 139 J\/kg-K (b) Since 425 K is above the Debye temperature, a good approximation for C<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To estimate the specific heat capacity of aluminum at 50 K and 425 K, we use different approximations based on the relationship between the given temperature and the Debye temperature (\u0398_D = 375 K).<\/p>\n\n\n\n

(a) Specific Heat at 50 K<\/strong><\/h3>\n\n\n\n

Since 50 K is much lower than \u0398_D, we use the low-temperature approximation given by the Debye model:<\/p>\n\n\n\n

[
C_v = A T^3
]<\/p>\n\n\n\n

where ( A ) is determined from the given value at 30 K:<\/p>\n\n\n\n

[
A = C_v \\frac{T^3}{T^3} = (0.81 \\text{ J\/mol-K}) \\times (30 K)^3
]<\/p>\n\n\n\n

[
A = 3.00 \\times 10^{-5} \\text{ J\/mol-K}^4
]<\/p>\n\n\n\n

Now, we compute ( C_v ) at 50 K:<\/p>\n\n\n\n

[
C_v = A T^3 = (3.00 \\times 10^{-5} \\text{ J\/mol-K}^4) \\times (50 K)^3
]<\/p>\n\n\n\n

[
C_v = 3.75 \\text{ J\/mol-K}
]<\/p>\n\n\n\n

The specific heat in J\/kg-K is obtained by dividing by the molar mass of aluminum (26.98 g\/mol):<\/p>\n\n\n\n

[
c_v = \\left( 3.75 \\text{ J\/mol-K} \\right) \\times \\left( \\frac{1 \\text{ mol}}{26.98 \\text{ g}} \\right) \\times \\left( 1000 \\text{ g\/kg} \\right)
]<\/p>\n\n\n\n

[
c_v = 139 \\text{ J\/kg-K}
]<\/p>\n\n\n\n

(b) Specific Heat at 425 K<\/strong><\/h3>\n\n\n\n

For temperatures above \u0398_D, the heat capacity follows the classical Dulong-Petit law:<\/p>\n\n\n\n

[
C_v \\approx 3R
]<\/p>\n\n\n\n

where ( R = 8.314 \\text{ J\/mol-K} ), so:<\/p>\n\n\n\n

[
C_v = 3 \\times 8.314 = 24.94 \\text{ J\/mol-K}
]<\/p>\n\n\n\n

The specific heat in J\/kg-K:<\/p>\n\n\n\n

[
c_v = \\left( 24.94 \\text{ J\/mol-K} \\right) \\times \\left( \\frac{1 \\text{ mol}}{26.98 \\text{ g}} \\right) \\times \\left( 1000 \\text{ g\/kg} \\right)
]<\/p>\n\n\n\n

[
c_v = 924 \\text{ J\/kg-K}
]<\/p>\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n
    \n
  • At low temperatures (T << \u0398_D)<\/strong>, the Debye model predicts that heat capacity follows the ( T^3 ) dependence.<\/li>\n\n\n\n
  • At high temperatures (T > \u0398_D)<\/strong>, atomic vibrations follow classical behavior, and the heat capacity approaches Dulong-Petit\u2019s law<\/strong> with a nearly constant value of 3R<\/strong>.<\/li>\n\n\n\n
  • Using these approximations, we find that at 50 K<\/strong>, ( c_v = 139 ) J\/kg-K, and at 425 K<\/strong>, ( c_v = 924 ) J\/kg-K.<\/li>\n<\/ul>\n\n\n\n

    Now, I will generate an image illustrating the temperature dependence of specific heat.<\/p>\n\n\n\n

    Here is the scientific diagram illustrating the specific heat capacity of aluminum as a function of temperature. The graph follows the Debye model at low temperatures (T\u00b3 dependence) and levels off at high temperatures according to the Dulong-Petit law (3R). Key points such as 30K, 50K, 375K (Debye temperature), and 425K are marked. Let me know if you need any modifications!<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    17.4 For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J\/mol-K and the Debye temperature is 375 K. Estimate the specific heat (a) at 50 K and (b) at 425 K. Solution (a) For aluminum, Cv at 50 K may be approximated by Equation 17.2, since this temperature is significantly […]<\/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-194040","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194040","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=194040"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194040\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194040"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194040"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}