A boiler with a steel bottom 15 mm thick rests on a hot stove; the area of the bottom of the boiler is 0.15 m^2. The water inside the boiler is at 100°C and 0.75 kg evaporates every minute.Calculate the temperature of the lower surface of the boiler, where it is in contact with the stove.
It will be important to look up what value your textbook gives for the thermal conductivity of steel. I'm going to use 50 W/(m K), but I found different values in different places. The latent heat of evaporation for water seems to be more of an agreed value: 2260 kJ/kg The heat required to evaporate 0.75 kg per minute is 1695 kJ/minute = 28.25 kW The heat transferred through the bottom of the boiler is k A delta-T / d where k = 50 W/(m K) A = 0.15 m^2 d = 0.015 m and delta-T is to be calculated. 28,250 W = 50 W/(m K) times 10 m times delta-T delta-T = 56.5 K I guess the temperature of the lower surface of the boiler is 156.5 Celsius. Check my arith.
