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 \(27^\circ\mathrm{C}\) to \(77^\circ\mathrm{C}\)? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).
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 \(27^\circ\mathrm{C}\) to \(77^\circ\mathrm{C}\)? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).
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Heat required \(=ms\Delta T\); net heating power \(=1000-160=840\) W.
Updated On: Nov 11, 2026
- \(8\) min \(20\) s
- \(7\) min
- \(14\) min
- \(6\) min \(2\) s
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The Correct Option is A
Solution and Explanation
Mass of water: 2 L = 2 kg. Heat needed: \(Q=ms\Delta T=2\times4200\times(77-27)=4.2\times10^5\) J.
Net power supplied to water: \(1000-160=840\) W.
Time: \(t=\dfrac{4.2\times10^5}{840}=500\) s \(=8\) min \(20\) s.
Answer: (1) 8 min 20 s
Net power supplied to water: \(1000-160=840\) W.
Time: \(t=\dfrac{4.2\times10^5}{840}=500\) s \(=8\) min \(20\) s.
Answer: (1) 8 min 20 s
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