Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In…

Question

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27oC to 77oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

MARKS Content Team · Accepted Answer

Correct answer: (A) 8 min 20 s. P = 1 0 0 0 watt ∴ P_in=1000-160 J/s=840 J/s Now, (d Q)/dt = ms dT/dt ∴ 840=msdT/dt ∴ dt = ms/(8 4 0) dT Integrating both sides we have; t = ms/(8 4 0) ∫ dT = (2 × 1 0 0 0 × 1 0^(- 6) × 1 0 0 0 × 4.2 × 1 0 0 0)/(8 4 0) × 5 0 = 5 0 0 sec = 8 min 2 0 sec

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27^oC to 77^oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

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