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InterviewSolution
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A resistor of resistance 200 Ω carries a current of 2 A for 10 minutes. Assuming that almost all the heat produced in the resistor is transferred to water (mass = 5 kg, specific heat capacity = 1 kcal/kg), and the work done by the water against the external pressure during the expansion of water can be ignored, find the rise in the temperature of the water. (J = 4186 J/cal) |
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Answer» Data : I = 2 A, R = 200 Ω, t = 10 min = 10 × 60 s = 600 s, M = 5 kg, S = (1 kcal/kg) (4186 J/kcal) = 4186 J/kg Q = ∆ U + W = MS ∆T + W ≃ MS ∆T ignoring W.. Also, Q = I2 Rt ∴ I2 RT = MS∆T ∴ The rise in the temperature of water = ∆T = \(\frac{I^2Rt}{MS}\) = \(\frac{(2)^2(200)(600)}{(5)(4186)}\)°C = 22.93°C |
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