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प्रश्न
A refrigerator converts 100g of water at 20℃ to ice at – 10℃ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J kg-1 K-1, specific latent heat of ice is 336 J g-1 and the specific heat capacity of ice is 2.1 J kg-1 K-1.
उत्तर
Amount of heat released when 100g of water cools from 20°C to 0°C = 100 × 20 × 4.2 = 8400J.
Amount of heat released when 100g of water converts into ice at 0°C = 100 × 336 = 33600J.
Amount of heat released when 100g of ice cools from 0°C to -10°C = 100 × 10 × 2.1 = 2100J.
Total amount of heat = 8400 + 33600 + 2100
= 44100J.
Time taken = 73.5 min
= 4410 s.
Average rate of heat extraction (power)
`"p" = "E"/"t"`
= `44100/4410`
= 10 J/s
= 10 W
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