Question

The heat of fusion of water is 79. 9 cal/g, the heat of vaporization of water is 540 cal/g, the specifc heat of ice is 0. 492 cal/deg/g, the specific heat of steam is 0. 488 cal/deg/g and the specific heat of liquid water is 1. 00 cak/deg/g. How much heat would be needed to convert 10. 51 g of ice at -30 °C to steam at 126 °C?

Answer 1

ΔQ = S M ΔT where S is the specific heat

ΔQ = 10.51 [.492 * (0 - (-30)) + 79.9 + 1.00 (100 - 0) + 540 + .488 (126 - 100)] = 10.51 * (.492 * 30 + 79.9 + 100 + 540 + .488 * 26)

Adding the heat required for each individual process

ΔQ = 10.51 * (14.8 + 79.9 + 100 + 540 + 12.7)

ΔQ = 10.51 gm * 747 cal/gm = 7860 cal