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Question
A heater supplies heat at the rate of 800 J/s. Find the time required to convert 50g of ice at -20°C into superheated steam at 140°C.
(Given: Specific heat of ice = 2.1 J/g, latent heat of ice = 340 J/gm, latent heat of steam = 2240 J/gm, specific heat of steam = 2.1 J/gm and specific heat of water = 4.2 J/gm.)
Solution
(i) Heat required to raise the temperature of ice from - 20°C to 0°C
= m × s × t = 50 × 2.1 × 20 = 2100 J
(ii) Heat required to melt ice at 0°C = mL = 50 × 340 = 1700 J
(iii) Heat required to heat water from °C to 100°C
= m × s × t = 50 × 4.2 × 100 = 21000 J
(iv) Heat required to convert water into steam at 100 C
= m × L = 50 × 2240 = 124000 J
(v) Heat required to heat steam from 100 C to 140 C
= m × s × t = 50 × 2.1 × 40 = 4200 J
Total quantity of heat = (2100 + 17000 + 21000 + 112000 + 4200) J
= 156300 J
If t is the time taken by heater, in second.
Thewn 800 × t = 156300 J
∴ t = `156300/800`s = 195.375 s = 195.4s
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