Question

What is the quantity of heat (in kJ) associated with cooling 187.1 g of water from 25.60 °C to ice at -10.70 °C?

a) -465.7 kJ
b) -348.9 kJ
c) -294.2 kJ
d) -182.6 kJ

Answer 1

The quantity of heat associated with cooling water can be calculated using the specific heat capacity of water.

Step-by-step explanation:

To calculate the quantity of heat associated with cooling water, we need to use the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. We can use the formula: q = m * c * ΔT, where q is the quantity of heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given: mass of water = 187.1 g, initial temperature = 25.60°C, final temperature = -10.70°C.

Using the formula, q = 187.1 g * 4.18 J/g°C * (-10.70°C - 25.60°C).

Converting the result from joules to kilojoules, we get q = -294.2 kJ. Therefore, the correct answer is c) -294.2 kJ.