P=1000 watt
∴Pin=(1000−160)J/s=840 J/s
Now, dtdQ=msdtdT
∴840=msdtdT
∴dt=840msdT
Integrating both sides we have;
t=840ms∫dT
=840(2×1000×10−6×1000)×4.2×1000×50
=500sec
=8min20sec
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).
Held on 9 Apr 2014 · Verified 6 Jul 2026.
8 min 20 s
7 min
14 min
6 min 2 s
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