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प्रश्न
A refrigerator converts 100 g of water at 20°C to ice at -10°C in 73.5 minutes. Calculate the average rate of heat extraction from water in watt. (Sp. heat of ice= 2100 J/kgK, sp. heat of water = 4200 J/kgK, Sp. latent heat of ice = 336000 J/kg.)
उत्तर
Heat given out during the following three stages:
1. Cooling water from 20°C tD 0°C = mC1θ1 = 100 x 4.2 x 20 = 8400 J
2. Water at 0°C freezes tD form ice at 0°C = m x L = 100 x 336 = 33600 J
3. Cooling of ice at 0°C tD -10°C =mCzθz = 100 x 2 .1 x 10 = 2100 J
Total quantity of heat given out= 44100 J
Rate of heat extraction in watts= `44100/(73.5 xx 60)` =10 W
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