Four ice cubes at exactly 0 ∘C with a total mass of 54.0 g are combined with 140 g of water at 90 ∘C in an insulated container. (ΔH∘fus =6.02 kJ/mol , cwater =4.18J/g⋅∘C ) If no heat is lost to the surroundings, what is the final temperature of the mixture?
The Correct Answer and Explanation is:
To solve this problem, we need to consider the energy balance in the system, where heat lost by the hot water is equal to the heat gained by the ice cubes as they melt and warm up to the final temperature.
Step 1: Heat required to melt the ice cubes
We begin by calculating how much heat is required to melt the ice cubes. The heat required to melt ice is given by the formula:Qmelt=n⋅ΔHfusQ_{\text{melt}} = n \cdot \Delta H_{\text{fus}}Qmelt=n⋅ΔHfus
where:
- nnn is the number of moles of ice,
- ΔHfus\Delta H_{\text{fus}}ΔHfus is the molar heat of fusion (6.02 kJ/mol).
First, we calculate the number of moles of ice:n=miceMice=54.0 g18.0 g/mol=3.00 moln = \frac{m_{\text{ice}}}{M_{\text{ice}}} = \frac{54.0 \, \text{g}}{18.0 \, \text{g/mol}} = 3.00 \, \text{mol}n=Micemice=18.0g/mol54.0g=3.00mol
So, the heat required to melt the ice is:Qmelt=3.00 mol×6.02 kJ/mol=18.06 kJ=18060 JQ_{\text{melt}} = 3.00 \, \text{mol} \times 6.02 \, \text{kJ/mol} = 18.06 \, \text{kJ} = 18060 \, \text{J}Qmelt=3.00mol×6.02kJ/mol=18.06kJ=18060J
Step 2: Heat required to warm the melted water
After the ice melts, it will warm up from 0 ∘C to the final temperature TfT_fTf. The heat required for this is:Qwarm=mwater⋅cwater⋅ΔTQ_{\text{warm}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta TQwarm=mwater⋅cwater⋅ΔT
where:
- mwater=54.0 gm_{\text{water}} = 54.0 \, \text{g}mwater=54.0g (mass of the melted ice),
- cwater=4.18 J/g⋅∘Cc_{\text{water}} = 4.18 \, \text{J/g⋅∘C}cwater=4.18J/g⋅∘C,
- ΔT=Tf−0 ∘C\Delta T = T_f – 0 \, \text{∘C}ΔT=Tf−0∘C.
Thus, the heat required to warm the melted water is:Qwarm=54.0 g×4.18 J/g⋅∘C×Tf=225.72 Tf JQ_{\text{warm}} = 54.0 \, \text{g} \times 4.18 \, \text{J/g⋅∘C} \times T_f = 225.72 \, T_f \, \text{J}Qwarm=54.0g×4.18J/g⋅∘C×Tf=225.72TfJ
Step 3: Heat lost by the hot water
Now, the hot water will cool down from 90 ∘C to the final temperature TfT_fTf. The heat lost by the hot water is:Qlost=mhot water⋅cwater⋅ΔTQ_{\text{lost}} = m_{\text{hot water}} \cdot c_{\text{water}} \cdot \Delta TQlost=mhot water⋅cwater⋅ΔT
where:
- mhot water=140 gm_{\text{hot water}} = 140 \, \text{g}mhot water=140g,
- cwater=4.18 J/g⋅∘Cc_{\text{water}} = 4.18 \, \text{J/g⋅∘C}cwater=4.18J/g⋅∘C,
- ΔT=90 ∘C−Tf\Delta T = 90 \, \text{∘C} – T_fΔT=90∘C−Tf.
Thus, the heat lost by the hot water is:Qlost=140 g×4.18 J/g⋅∘C×(90−Tf)=585.2 (90−Tf) JQ_{\text{lost}} = 140 \, \text{g} \times 4.18 \, \text{J/g⋅∘C} \times (90 – T_f) = 585.2 \, (90 – T_f) \, \text{J}Qlost=140g×4.18J/g⋅∘C×(90−Tf)=585.2(90−Tf)J
Step 4: Energy balance equation
According to the principle of conservation of energy, the heat lost by the hot water equals the heat gained by the ice cubes. So, we can write the energy balance as:Qlost=Qmelt+QwarmQ_{\text{lost}} = Q_{\text{melt}} + Q_{\text{warm}}Qlost=Qmelt+Qwarm
Substituting the expressions we derived:585.2 (90−Tf)=18060+225.72 Tf585.2 \, (90 – T_f) = 18060 + 225.72 \, T_f585.2(90−Tf)=18060+225.72Tf
Expanding both sides:585.2×90−585.2 Tf=18060+225.72 Tf585.2 \times 90 – 585.2 \, T_f = 18060 + 225.72 \, T_f585.2×90−585.2Tf=18060+225.72Tf52668−585.2 Tf=18060+225.72 Tf52668 – 585.2 \, T_f = 18060 + 225.72 \, T_f52668−585.2Tf=18060+225.72Tf
Rearranging terms to solve for TfT_fTf:52668−18060=585.2 Tf+225.72 Tf52668 – 18060 = 585.2 \, T_f + 225.72 \, T_f52668−18060=585.2Tf+225.72Tf34608=810.92 Tf34608 = 810.92 \, T_f34608=810.92TfTf=34608810.92≈42.7 ∘CT_f = \frac{34608}{810.92} \approx 42.7 \, \text{∘C}Tf=810.9234608≈42.7∘C
Final Temperature
The final temperature of the mixture is approximately 42.7 ∘C.
