How many kilojoules are needed to raise the temperature of 325 grams of water from 15.0 degrees c to 77.0 degrees c?

Given:
m = 325 g, the mass of water
ΔT = 77 - 15 = 62 °C, the change in temperature

Note that
c = 4.184 J/(g-°C), the specific heat of water.

Use the formula
Q = m*c*ΔT
where Q is the heat required for the change in temperature.

Q = (325 g)(4.184 J/g-°C)*(62 °C) = 8.43 x 10⁴ J
That is,
Q = 84.3 x 10³ J = 84.3 kJ

  84.3 kJ

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