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$A$ 10 W electric heater is used to heat a container filled with $0.5 \mathrm{kg}$ of water. It is found that the temperature of water and the container rises by $3^{\circ} \mathrm{K}$ in $15 \mathrm{min}$. The container is then emptied, dried and filled with $2 \mathrm{kg}$ of oil. The same heater now raises the temperature of container-oil system by $2^{\circ} \mathrm{K}$ in $20 \mathrm{min}$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \mathrm{d} \mathrm{kg}^{-1} \mathrm{K}^{-1}$, the specific heat of oil in the same unit is equal to
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Verified Answer
The correct answer is:
$2.55 \times 10^{3}$
$m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=$ Work done
$m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=P_{1} t_{1}$
where $m_{1}=0.5 \mathrm{kg}$
Specific heat $s_{1}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}$
$$
\begin{array}{l}
\Delta t=\Delta t_{1}=\Delta t_{2}=3 \mathrm{K} \\
P_{1}=P_{2}=10 \mathrm{W} \\
t_{1}=15 \times 60=900 \mathrm{s}
\end{array}
$$
$s_{2}=$ Specific heat capacity of container
$$
0.5 \times 4200 \times(3-0)+m_{2} \mathrm{s}_{2} \times(3-0)
$$
$$
\begin{aligned}
&=10 \times 15 \times 60 \\
2100 \times 3+m_{2} s_{2} \times 3 &=9000 \\
m_{2} s_{2} &=\frac{9000-6300}{3} \\
m_{2} s_{2} &=900
\end{aligned}
$$
Similarly, in case of oil
$$
m_{1} s_{0} \Delta t+m_{2} s_{2} \Delta t=P_{2} t_{2}
$$
where $s_{0}=$ specific heat capacity of oil
$$
\begin{aligned}
P_{1} &=P_{2}=10 \mathrm{W} \\
2 \times s_{0} \times 2+900 \times 2 &=10 \times 20 \times 60 \\
4 s_{0}+1800 &=12000 \\
4 s_{0} &=12000-1800 \\
s_{0} &=\frac{10200}{4}=2550 \\
&=2.55 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1} \mathrm{k}^{-1}
\end{aligned}
$$
$m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=P_{1} t_{1}$
where $m_{1}=0.5 \mathrm{kg}$
Specific heat $s_{1}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}$
$$
\begin{array}{l}
\Delta t=\Delta t_{1}=\Delta t_{2}=3 \mathrm{K} \\
P_{1}=P_{2}=10 \mathrm{W} \\
t_{1}=15 \times 60=900 \mathrm{s}
\end{array}
$$
$s_{2}=$ Specific heat capacity of container
$$
0.5 \times 4200 \times(3-0)+m_{2} \mathrm{s}_{2} \times(3-0)
$$
$$
\begin{aligned}
&=10 \times 15 \times 60 \\
2100 \times 3+m_{2} s_{2} \times 3 &=9000 \\
m_{2} s_{2} &=\frac{9000-6300}{3} \\
m_{2} s_{2} &=900
\end{aligned}
$$
Similarly, in case of oil
$$
m_{1} s_{0} \Delta t+m_{2} s_{2} \Delta t=P_{2} t_{2}
$$
where $s_{0}=$ specific heat capacity of oil
$$
\begin{aligned}
P_{1} &=P_{2}=10 \mathrm{W} \\
2 \times s_{0} \times 2+900 \times 2 &=10 \times 20 \times 60 \\
4 s_{0}+1800 &=12000 \\
4 s_{0} &=12000-1800 \\
s_{0} &=\frac{10200}{4}=2550 \\
&=2.55 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1} \mathrm{k}^{-1}
\end{aligned}
$$
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