Quantity of Heat Calculation for Water Cooling Process

What is the quantity of heat (in kJ) associated with cooling 121.9 g of water from 25.60 °C to ice at -10.70 °C?

A) 12.19 kJ

B) 24.38 kJ

C) 36.57 kJ

D) 48.76 kJ

Final answer:

To calculate the quantity of heat associated with cooling 121.9 g of water from 25.60 °C to ice at -10.70 °C, we can use the equation Q = mcΔT. First, we calculate the heat required to cool the water from 25.60 °C to 0 °C, and then the heat required to freeze the water from 0 °C to -10.70 °C. Adding these two values together gives a total quantity of heat of approximately 36.57 kJ.

Explanation:

To calculate the quantity of heat associated with cooling 121.9 g of water, we can use the equation Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to calculate the heat required to cool the water from 25.60 °C to 0 °C, using the specific heat capacity of water (4.184 J/g °C). Then, we need to calculate the heat required to freeze the water from 0 °C to -10.70 °C, using the heat of fusion of water (79.8 cal/g or 334 kJ/kg).

Adding the two values together, we get a total quantity of heat of approximately 36.57 kJ. Therefore, the answer is C) 36.57 kJ.