How much heat (in kJ) is required to increase the temperature of a 78.0 g sample

Question

How much heat (in kJ) is required to increase the temperature of a 78.0 g sample of water from 84.0 to 127.0 °C?

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Explanation / Answer

m= mass = 78.0 g

dT = 127 - 84 = 43 oC

Cp = 4.184 J / g oC

Q = m Cp dT

Q = 78 x 4.184 x 43

Q = 14033 J

Q = 14.0 kJ

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