Answer:
The minimum quantity of steam required per pound of ice is -22.5 BTU/lbm
Explanation:
Q = mCT
Q/m = CT
Q/m is the quantity of steam required per pound of ice
C is the heat capacity of ice = 0.45 BTU/lbm°F
T is the temperature of ice = -50 °F
Q/m = 0.45 BTU/lbm°F × -50 °F = -22.5 BTU/lbm
