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How much heat (in joules) is required to raise the temperature of 36.2 kg of water from 17 °C to 93 °C?

Express your answer using two significant figures.

Answer :

Final answer:

The heat required to raise the temperature of 36.2 kg of water from 17 °C to 93 °C is approximately 1.1 x 10^7 Joules.

Explanation:

The subject of this question relates to the concept of 'specific heat', which is a part of Physics. In this scenario, we are asked to calculate the amount of heat required to raise the temperature of a certain mass of water. The formula used to calculate the heat needed is: q = mcΔT, where 'q' represents heat, 'm' is the mass of the substance, 'c' is the specific heat, and 'ΔT' is the change in temperature.

To find the heat in joules, we substitute the given values into our formula: The mass (m) is 36.2 kg (or 36200g, because 1kg = 1000g), the specific heat (c) of water is 4.184 J/g°C, and the change in temperature (ΔT) is 93°C - 17°C = 76°C. Therefore, q = 36200g * 4.184 J/g°C * 76°C = 11459689.6 J. When rounded off to two significant figures, it becomes 1.1 x 10^7 J.

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