Question

Water of volume $2L$ in a closed container is heated with a coil of $1kW$. 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 ${27}^{o}C$ to ${77}^{o}C$? (Specific heat of water is $4.2kJ/kg$ and that of the container is negligible)

• A. $8min20s$
• B. $6min2s$
• C. $7min$
• D. $14min$

Answer 1

The correct option is A $8min20s$

Heat energy required to raise the temp of water to ${77}^{o}C$
$Q_w=2\times 4.2\times (77-27)\times 1000=420000J$
Heat lost for time t is
$Q_L=160t$

In time t, total energy supplied is 1000t. Therefore,
$1000t=Q_w+Q_L$
Or,
$t=\frac{420000}{840}=500s=8min20s$