Heat and Thermodynamics 1 Question 12
12. Water of volume $2 L$ in a container is heated with a coil of $1 kW$ at $27^{\circ} C$. The lid of the container is open and energy dissipates at rate of $160 J / s$. In how much time temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ ?
(2005, 2M)
[Specific heat of water is $4.2 kJ / kg$ ]
(a) $8 min 20 s$
(b) $6 \min 2 s$
(c) $7 min$
(d) $14 min$
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Answer:
Correct Answer: 12. (a)
Solution:
- Energy gained by water (in $1 s$ )
$=$ energy supplied - energy lost $=1000 J-160 J=840 J$
Total heat required to raise the temperature of water from $27^{\circ} C$ to $77^{\circ} C$ is $m s \Delta \theta$.
Hence, the required time
$$ \begin{aligned} t & =\frac{m s \Delta \theta}{\text { rate by which energy is gained by water }} \\ & =\frac{(2)\left(4.2 \times 10^{3}\right)(50)}{840} \\ & =500 s=8 min 20 s \end{aligned} $$
