CalorimetryMedium
Question
A refrigerator converts 100 g of water at 25oC into ice at - 10oC in one hour and 50 minutes. The quantity of heat removed per minute is : - (Specific heat of ice = 0.5 cal/goC, latent heat of fusion = 80 cal/g )
Options
A.50 cal
B.100 cal
C.200 cal
D.75 cal
Solution
Heat removed in cooling water from 25oC to 0oC = 100 × 1 × 25 = 2500 cal
Heat removed in converting water into ice at 0oC = 10 × 80 = 8000 cal
Heat removed in cooling ice from
0o to - 15oC = 100 × 0.5 × 10 = 500 cal
Total heat removed in 1 hr 50 min = 2500 + 8000 + 500 = 11000 cal
Heat removed per minute =
= 100 cal / min
Heat removed in converting water into ice at 0oC = 10 × 80 = 8000 cal
Heat removed in cooling ice from
0o to - 15oC = 100 × 0.5 × 10 = 500 cal
Total heat removed in 1 hr 50 min = 2500 + 8000 + 500 = 11000 cal
Heat removed per minute =
= 100 cal / minCreate a free account to view solution
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