Question

A 10 watt 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 $3^o$ K in 15 minutes. The contain r is then emptied, dried and filled with 2 kg of oil. The same heater now raises the temperature of container-oil system by $2^o$K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 $Jk{g}^{-1}{K}^{-1}$ , the specific heat of oil in the same unit is equal to

• A. $1.50\times {10}^3$
• B. $2.55\times {10}^3$
• C. $3.00\times {10}^3$
• D. $5.10\times {10}^3$

Answer 1

The correct option is B $2.55\times {10}^3$

Power of heater $P=10$ W

Let the specific heat of container be $S_c$ and that of oil be $S_o$.

Let mass of container be $m_c$.

For water-container system :

Given : $m_w=0.5$ kg $\Delta T=3^o$ K $t=15\times 60=900$ s $S_w=4200$ $Jk{g}^{-1}{K}^{-1}$

Heat given $Pt=m_w{S}_w\Delta T+m_c{S}_c\Delta T$

$\therefore$ $10\left(900\right)=0.5\left(4200\right)\left(3\right)+m_c{S}_c\left(3\right)$ $⟹m_c{S}_c=900$ $J{K}^{-1}$ ......(1)

For oil-container system :

Given : $m_o=2$ kg $\Delta T=2^o$ K $t=20\times 60=1200$ s

Heat given $Pt=m_o{S}_o\Delta T+m_c{S}_c\Delta T$

$\therefore$ $10\left(1200\right)=\left(2\right)S_o\left(2\right)+\left(900\right)\left(2\right)$

$⟹S_o=2.55\times {10}^3$ $Jk{g}^{-1}{K}^{-1}$