InterviewSolution
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InterviewSolution
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1 kg of water initially at a temperature 27°C is heated by a heater of power 1KW. If the lid is opened, heat is lost at a constant rate of 200J/S. Find the time required for water to attain a temperature of 80°C with the lid open. (Specific heat of water = 4.2kJ/ kg/k) |
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Answer» Amount of water = 1 kg Net heat supplied = (1000 – 200) = 800 W Change in temperature = ∆ T = (353 – 300) = 53K ∴ ∆ T = 53K Time required, t = H/R where H is the required energy and R is the rate of heat transfer H = mC ∆T = 1 × 4.2 × 103 × 53 = 2.226 × 105 J R = 8 × 102 J/s ∴ t = \(\frac{2.226 \times 10^5}{8\times 10^2}\) = 2.783 × 102 ∴ t ≈ 279 = 4 min 39 secs. |
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