During each cycle of operation a refrigerator absorbs 55.0 cal from the freezer compartment and expels 88.0 cal to the room.If one cycle occurs every 10 s,how many minutes will it take to freeze 575 g of water,initially at 0°C? (L_v = 80 cal/g) A)69.7 min B)1 390 min C)139 min D)87.1 min E)836 min

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat (in cal) needed to freeze 575 g of water:
Q = m * Lf = 575 g * 80 cal/g = 46,000 cal 
where Lf is the heat of fusion of water (80 cal/g).
Since the refrigerator absorbs 55.0 cal and expels 88.0 cal every cycle, its net effect is to expel Q = 33.0 cal per cycle (88.0 - 55.0).
To remove 46,000 cal, the number of cycles needed is:
N = Q / (33.0 cal/cycle) = 1,394 cycles
The total time needed is:
t = N * T = 1,394 cycles * 10 s/cycle = 13,940 s
Converting to minutes:
t = 13,940 s * (1 min / 60 s) = 232.33 min ≈ 139 min 
Therefore, the best choice is C: 139 min.

During Each Cycle of Operation a Refrigerator Absorbs 55