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An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 ℃ to 15.0 ℃ I 100 s. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity of the liquid. |
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Answer» Power of heater P = 600 W Mass of liquid m = 4.0 kg Change in temperature of liquid = (15 − 10)oC = 5oC(or 5 K) Time taken to raise its temperature = 100s Heat energy required to heat the liquid ΔQ = mcΔT And ΔQ = P × t = 600 × 100 = 60000J c = ΔQ/mΔT = 60000/4 x 5 = 3000 Jkg-1K-1 Heat capacity = c × m Heat capacity = 4 × 3000 = 1.2 × 104 J/K |
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