Two 20.0 g ice cubes at -14.0 °C-14.0 °C are placed into 295 g295 g of water at 25.0 °C.25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, ??f,Tf, of the water after all the ice melts.
heat capacity of H2O(s)H2O(s)
37.7 J/(mol·K)37.7 J/(mol·K)
heat capacity of H2O(l)H2O(l)
75.3 J/(mol·K)75.3 J/(mol·K)
enthalpy of fusion of H2OH2O
6.01 kJ/mol6.01 kJ/mol
??f=Tf=
°C
The Correct Answer and Explanation is :
The final temperature of the water after all the ice melts is approximately 13.2°C.
Explanation:
- Heat Required to Warm Ice from -14.0°C to 0°C
Ice must first be warmed up to 0°C before melting. Using the specific heat capacity of ice:
[
q_1 = n_{\text{ice}} \times C_{\text{ice}} \times \Delta T
]
where:
- ( n_{\text{ice}} = \frac{40.0 \text{ g}}{18.015 \text{ g/mol}} = 2.22 \text{ moles} )
- ( C_{\text{ice}} = 37.7 \text{ J/(mol·K)} )
- ( \Delta T = 0 – (-14) = 14 \text{ K} ) [
q_1 = 2.22 \times 37.7 \times 14 = 1170.9 \text{ J}
]
- Heat Required to Melt Ice at 0°C
The heat needed to melt the ice completely:
[
q_2 = n_{\text{ice}} \times \Delta H_{\text{fusion}}
]
where ( \Delta H_{\text{fusion}} = 6.01 \text{ kJ/mol} = 6010 \text{ J/mol} ). [
q_2 = 2.22 \times 6010 = 13342.2 \text{ J}
] Total heat required to warm and melt the ice:
[
q_{\text{total}} = q_1 + q_2 = 1170.9 + 13342.2 = 14513.1 \text{ J}
] - Heat Lost by Water
The water will release heat to the ice:
[
q_{\text{water}} = n_{\text{water}} \times C_{\text{water}} \times (T_{\text{initial, water}} – T_f)
]
where:
- ( n_{\text{water}} = \frac{295.0 \text{ g}}{18.015 \text{ g/mol}} = 16.38 \text{ moles} )
- ( C_{\text{water}} = 75.3 \text{ J/(mol·K)} )
- ( \Delta T = (25 – T_f) ) The total heat available from the water:
[
q_{\text{available}} = 16.38 \times 75.3 \times (25 – 0) = 30866.3 \text{ J}
] Since ( q_{\text{available}} > q_{\text{total}} ), all ice melts, and the remaining heat increases the final temperature.
- Solving for ( T_f )
The remaining heat after melting:
[
q_{\text{remaining}} = q_{\text{available}} – q_{\text{total}} = 30866.3 – 14513.1 = 16353.2 \text{ J}
] The final temperature is found using:
[
T_f = \frac{q_{\text{remaining}}}{n_{\text{water}} \times C_{\text{water}}}
] [
T_f = \frac{16353.2}{16.38 \times 75.3} = 13.2°C
]
Thus, after all the ice melts, the equilibrium temperature of the system is 13.2°C.
Now, I’ll generate an image to illustrate this process.
The bar chart above illustrates the heat transfer process during the interaction of ice and water:
- Heating Ice (-14°C to 0°C): The ice requires 1170.9 J to warm up to 0°C.
- Melting Ice at 0°C: The phase change from solid to liquid consumes 13,342.2 J.
- Cooling Water (25°C to Final Temperature): The water releases 16,353.2 J, bringing the final temperature of the system to 13.2°C.
This visualization helps to understand the energy exchange involved in reaching thermal equilibrium.