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).

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°C to 77°C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible). (A) 8 min 20 s (B) 6 min 2 s (C) 7 min (D) 14 min

Tardigrade Admin · Accepted Answer

1000 - 160 = 840 J/s t-840=2×4.2× 103× 50 t=(500/60)=8 min 20 s.

8 min 20 s

6 min 2 s

7 min

14 min

1000−160=840J/s t−840=2×4.2×103×50 t=60500​=8min20s.

$1000 - 160 = 840 J / s$
$t - 840 = 2 \times 4.2 \times 1 0^{3} \times 50$
$t = \frac{500}{60} = 8 min 20 s .$