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प्रश्न
A refrigerator converts 100 g of water at 20°C to ice at -10°C in 35 minutes. Calculate the average rate of heat extraction in terms of watts.
Given: Specific heat capacity of ice = 2.1 J g-1°C-1
Specific heat capacity of water = 4.2 J g-1°C-1
Specific latent heat of fusion of ice = 336 J g-1
उत्तर
Given that
Mass of water converted to ice = m = 100 g
Temperature of water tw = 20°C
Temperature of ice ti = –10°C
Total time t = 35 min = 2100 s
Specific heat capacity of ice = 2.1 J g-1°C-1
Specific heat capacity of water = 4.2 J g-1°C-1
Specific latent heat of fusion of ice = 336 J g-1
Amount of heat released when 100 g water cools from 20C to 0°C is
Q1 = mcΔT
= 100 x 4.2 x 20
= 8400 J
Amount of heat released when 100 g water converts to ice at 0°C is
Q2 = mL
= 100 x 336
= 33600 J
Amount of heat released when 100 g ice cools from 0°C to −10°C is
Q3 = mcΔT
= 100 x 2.1 x 10
= 2100 J
Hence, the total heat released is
Q = Q1 + Q2 + Q3
Q = 8400 + 33600 + 2100
Q = 44100 J
Therefore, the average rate of heat extraction is
P = `Q/t = 44100/2100` = 21 W
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