Q54 / 59 · KVPY 2013 (SA) · Physics
A $10.0 \mathrm{~W}$ electrical heater is used to heat a container filled with $0.5 \mathrm{~kg}$ of water. It is found that the temperature of the water and the container rise by $3 \mathrm{K}$ in $15 \mathrm{~minutes}$. The container is then emptied, dried, and filled with $2 \mathrm{~kg}$ of an oil. It is now observed that the same heater raises the temperature of the container-oil system by $2 \mathrm{K}$ in $20 \mathrm{~minutes}$. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is
- Correct answerA$2.5 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
- B$5.1 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
- C$3.0 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
- D$1.5 \times 10^{3} \mathrm{JK}^{-1} \mathrm{~kg}^{-1}$
Solution
$\mathrm{Pt}=\mathrm{m}_{\mathrm{w}} \mathrm{S}_{\mathrm{w}} \Delta \mathrm{T}+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} \Delta \mathrm{T}$ $10 \times 15 \times 60=0.5 \times 4200 \times 3+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} \times 3$ $9000=6300+\mathrm{m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}} 3$
$\mathrm{~m}_{\mathrm{c}} \mathrm{s}_{\mathrm{c}}=900 \mathrm{~J} / \mathrm{k}$.
Now, for oil
$10 \times 20 \times 60=2 \times \mathrm{S}_{0} \times 2+900 \times 2$
$12000-1800=4 \mathrm{~S}_{0}$
$\mathrm{~S}_{0}=\frac{10200}{4}=2.51 \times 10^{3} \mathrm{~J} / \mathrm{kg}-\mathrm{k}$
Hence, the correct answer is option: 'A'.