Question

A $10W$ electric heater is used to heat a container filled with $0.5kg$ of water. It is found that the temperature of water and the container rises by $3K$ in $15min$. The container is then emptied, dried and filled with $2kg$ of oil. The same heater now raises the temperature of container-oil system by $2K$ in $20min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200J{kg}^{-1}{K}^{-1}$, 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 $2.55\times {10}^3$

Let the mass and specific heat of the container be $m_c$ and $S_c$ respectively.

For water : $Q=m_w{S}_w\Delta T+m_c{S}_c\Delta T$

$\therefore$ $\frac{dQ}{dt}=P=\frac{(m_w{S}_w+m_c{S}_c)\Delta T}{t}$ ............(1)

Given : $t=15$ minutes $=900$ s $\Delta T=3K$

Equation (1), $10=\frac{(0.5\times 4200+m_c{S}_c)\times 3}{900}$

$⟹m_c{S}_c=900$ $J{K}^{-1}$

Let the specific heat of oil be $S_o$

For oil : $Q=m_o{S}_o\Delta T^{\prime }+m_c{S}_c\Delta T^{\prime }$

$\therefore$ $\frac{dQ}{dt}=P=\frac{(m_o{S}_o+m_c{S}_c)\Delta T^{\prime }}{t}$ .

Given :$t=20$ minutes $=1200$ s $\Delta T^{\prime }=2K$

$10=\frac{(2\times S_o+900)\times 2}{1200}$

$⟹S_o=2550$ $Jk{g}^{-1}{K}^{-1}$