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A $10.0 \,W$ electrical heater is used to heat a container filled with $0.5 \,kg$ of water.It is found that the temperature of the water and the container rose by $3 \,K$ in $15 \,min$. The container is then emptied, dried and filled with $2 \,kg$ of an oil. It is now observed that the same heater raises the temperature of the container-oil system by $2 \,K$ in $20 \,min$. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is ................ $ \times 10^3 \,JK ^{-1} kg ^{-1}$
$25$
$51$
$3.0$
$15$
Solution
(a)
Energy supplied by heater $=$ Heat absorbed by water + Heat absorbed by oil So, with water in container,
$P \Delta t=m_w s_w \Delta T+m_o s_0 \Delta T$
$\Rightarrow 10 \times 15 \times 60=0.5 \times 4200 \times 3+m_o s_o \times 3$
$\Rightarrow m_o s_o=900 \,J K ^{-1}$
$\text { Now with oil in container, }$
$P \Delta t=m_o s_o \Delta T+m_c s_c \Delta T$
$\Rightarrow 10 \times 20 \times 60=2 \times s_o \times 2+900 \times 2$
$\Rightarrow s_o=\frac{10200}{4}=25 \times 10^3 \,J K ^{-1} kg ^{-1}$
