How much heat (in kJ) is required to warm 10.0 g of ice, initially at −10.0 °C, to steam at 110.0 °C? The heat capacity of ice is 2.09 J/g · °C, and that of steam is 2.01 J/g °C.

4.719 kJ

Explanation:

Q = m(L1∆T + L2T2)

m = 10 g

L1 = 2.09 J/g.°C

L2 = 2.01 J/g.°C

T2 = 110°C

T1 = -10°C

∆T = T2 - T1 = 110-(-10) = 120°C

Q = 10(2.09×120 + 2.01×110) = 10(250.8 + 221.1) = 10(471.9) = 4719 J = 4719/1000 = 4.719 kJ

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