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 $3^{\circ }K$ 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 $2^{\circ }K$ in $20min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200Jk{g}^{-1}{K}^{-1}$, the specific heat of oil in the same unit is equal to

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

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 $m_1=0.5kg$
Specific heat $s_1=4200J/kg-K$
$\Delta t=\Delta t_1=\Delta t_2=3K$
$P_1=P_2=10W$
$t_1=15\times 60=900s$
$s_2=$ 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
$P_1=P_2=10W$
$2\times s_0\times 2+900\times 2=10\times 20\times 60$
$4s_0+1800=12000$
$4s_0=12000-1800$
$s_0=\frac{10200}{4}=2550$
$=2.55\times 10^{3}Jk{g}^{-1}{K}^{-1}$