The heat capacity of ice is 2.09 J/g°C and the heat of fusion of ice is 6.02 kJ/mol. An ice cube at −12.0°C is placed into a glass of water at room temperature.
Does the melting of the ice or the warming of the ice from −12.0°C play a larger role in cooling the liquid water? A. They contribute equally B. The warming of the ice from −12.0°C C. The melting of the ice D. Cannot determine from the information given
The Correct Answer and Explanation is:
To determine whether the warming of the ice from −12.0°C or the melting of the ice contributes more to cooling the liquid water, we need to consider the energy changes involved in both processes.
Step 1: Calculate the energy required to warm the ice.
- Heat Capacity of Ice (C): 2.09 J/g°C
- Mass of Ice (m): Assuming 1 g for simplicity (this can be scaled later).
- Temperature Change (ΔT): From -12.0°C to 0°C, so ΔT = 0 – (-12.0) = 12.0°C.
Using the formula for heat transfer:
[ q = m \times C \times \Delta T ]
Plugging in the values:
[ q_{warming} = 1 \, \text{g} \times 2.09 \, \text{J/g°C} \times 12.0 \, \text{°C} = 25.08 \, \text{J} ]
Step 2: Calculate the energy required to melt the ice.
- Heat of Fusion (ΔH_fusion): 6.02 kJ/mol = 6020 J/mol.
- Molar Mass of Water: 18.02 g/mol, so for 1 g of ice:
[ \text{Moles of ice} = \frac{1 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0555 \, \text{mol} ]
Now calculate the energy required to melt 1 g of ice:
[ q_{melting} = 0.0555 \, \text{mol} \times 6020 \, \text{J/mol} \approx 334.11 \, \text{J} ]
Conclusion
Now, comparing the two values:
- Warming of the ice: 25.08 J
- Melting of the ice: 334.11 J
The melting of the ice requires significantly more energy than the warming of the ice. Therefore, the process of melting plays a much larger role in cooling the liquid water than the warming of the ice does.
Thus, the correct answer is C. The melting of the ice.