Question

Heat is supplied to 20lb of ice at 0°F at the rate of 160Btu/sec. How long will it take to convert the ice to steam at 213°C?

Answer 1

The correct problem:

Heat is supplied to 20 lb of ice at 0 °F at the rate of 160 Btu/sec. How long will it take to convert the ice to steam at 213 °F?

A key rule to remember is that 0.5 Btu of heat is required to raise 1 pound of ice 1° F when the temperature is below 32 °F; and 0.5 Btu of heat is required to raise 1 pound of steam 1°F above the temperature of 212 °F.

Latent heat of melting: 144 Btu/lb

Latent heat of evaporation: 970.4 Btu/lb

$Q_1 = 32 \times 20 \times 0.5 = 320 \;Btu$ (from 0 °F to 32 °F)

$Q_2 = 180 \times 20 \times 1 = 3600 \;Btu$ (from 32 °F to 212 °F)

$Q_3 = 1 \times 20 \times 0.5 = 10 \; Btu$ (from 212 °F to 213 °F)

$Q_3 = 20 \times 144 = 2880 \;Btu$ (heat of melting)

$Q_4 = 20 \times 970.4 = 19408 \;Btu$ (heat of evaporation)

Total heat:

$Q_T = 320 + 3600 + 10 + 2880 + 19408 = 28218 \;Btu$

$Time = \frac{Total \;heat }{Rate}$

$Time = \frac{28218 }{160} = 163.86 \;sec$