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← JEE Main 2014 · 9 Apr · Shift 1
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JEE Main 2014 · 9 Apr · Shift 1, Physics question 11

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°C to 77°C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

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Answer and worked solution

Official answer: A

Heat needed: 2×4200×50=4.2×1052\times4200\times50=4.2\times10^5 J. The net power is 1000−160=8401000-160=840 W, so
t=4.2×105840=500 s=8 min 20 s.t=\frac{4.2\times10^5}{840}=500\text{ s}=8\text{ min }20\text{ s}.