Question

A $10\, W$ electric heater is used to heat $a$ container filled with $0.5\, kg$ of water. It is found that the temperature of water and the container rises by $3^{\circ}\, K$ in $15 \,min .$ The container is then emptied, dried and filled with $2 \,kg$ of oil. The same heater now raises the temperature of container-oil system by $2^{\circ} \,K$ in $20 \,min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200\, J\, kg ^{-1} \,K ^{-1}$, the specific heat of oil in the same unit is equal to

• A. $1.50×10^3$
• B. $2.5×10^3$
• C. $3.00×10^3$
• D. $5.10×10^3$

Answer 1

Answer: (B)

$m_{1} \,s_{1} \,\Delta t+m_{2} \, s_{2} \, \Delta t=$ Work done
$m_{1} \,s_{1} \, \Delta t+m_{2} \,s_{2} \,\Delta t=P_{1} t_{1}$
where $m_{1}=0.5 \, kg$
Specific heat $s_{1}=4200 \,J / kg - K$
$\Delta t=\Delta t_{1}=\Delta t_{2}=3 \,K$
$P_{1}=P_{2}=10 \, W$
$t_{1}=15 \times 60=900 s$
$s_{2}=$ Specific heat capacity of container
$0.5 \times 4200 \times(3-0) +m_{2} s_{2} \times(3-0) $
$=10 \times 15 \times 60$
$2100 \times 3+m_{2} S_{2} \times 3 =9000$
$m_{2} S_{2} =\frac{9000-6300}{3} $
$m_{2} s_{2} =900$
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
$P_{1}=P_{2}=10\, W$
$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} \,J \,kg ^{-1}\, K ^{-1} $