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प्रश्न
1 kg of water is contained in a 1.25 kW kettle. Assuming specific heat capacity of water = 4.2 J/g °C and specific latent heat of vaporization = 2260 J/g, calculate:
(i) the time taken for the temperature of water to rise from 25°C to its boiling point,
(ii) the mass of water which evaporates per minute from the boiling water.
उत्तर
(i) Heat required for the temperature of 1 kg (= 1000 g) of water to rise from 25°C to its boilling point (i.e. 100°C)
= Mass × Specific heat capacity × Rise in temperature
= 1000 × 4.2 × (100 - 25) = 31500 J
Power supplied by the kettle = 1.25 kW = 1.25 × 1000 W = 1250 W
Since Power = `"Energy"/"Time"`
`therefore 1250 = 31500/"t"`
Hence time taken t = `315000/1250 = 252"s"` (or 4 min 12 s)
(ii) Energy supplied by the kettle in 1 minute (= 60s) = Power × Time
= 1250 × 60 = 75000 J
Heat required for boiling water to evaporate
= Mass × Specific latent heat of vaporization
= m × 2260 = 2260 mJ
Thus 2260 m = 75000
or mass of water evaporated per minute, m = `75000/2260 = 33.18 "g"`
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