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How much heat in kJ is absorbed by 652.3 g of water in order doe the temperature to increase from 25.0 c to 32.50 C?

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Final answer:

The heat absorbed by 652.3 g of water for its temperature to increase from 25.0°C to 32.5°C is 20.43 kJ. This is calculated using the specific heat capacity for water and the temperature change.

Step-by-step explanation:

To calculate the heat absorbed by water when the temperature increases, we use the formula q = m × c × ΔT. Here, q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.184 J/g°C.

For 652.3 g of water to increase in temperature from 25.0°C to 32.5°C, calculate the temperature change (ΔT) which is 32.5°C - 25.0°C = 7.5°C.

The heat absorbed (q) is then calculated as follows:
q = 652.3 g × 4.184 J/g°C × 7.5°C, which equals 20430.3 J or 20.43 kJ since 1 kJ = 1000 J.

Thus, the water absorbs 20.43 kJ of heat.

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