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"How much heat (in joules) is required to raise the temperature

of 19.5 kg of water from 15°C to 92°C?

1 Answer

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Final answer:

To raise the temperature of 19.5 kg of water from 15°C to 92°C, 6293532 joules of heat are required. This calculation uses the specific heat capacity of water, which is 4.184 J/g°C, and the mass-temperature change relationship from the formula q = mcΔT.

Step-by-step explanation:

To calculate the amount of heat needed to raise the temperature of 19.5 kg of water from 15°C to 92°C, we need to use the specific heat capacity of water and the formula q = mcΔT, where q is the heat in joules, m is the mass of the water in grams, c is the specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of water is 4.184 J/g°C. The mass (m) needs to be in grams, so we convert 19.5 kg to 19500 g. The temperature change (ΔT) is 92°C - 15°C = 77°C. Now, we can use the formula to find the heat required:

q = 19500 g * 4.184 J/g°C * 77°C
q = 6293532 J

Therefore, the amount of heat required to raise the temperature of 19.5 kg of water from 15°C to 92°C is 6293532 joules.

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