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 container rises by 3K 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 2K in  20 min. Assume there is no heat loss in the process and the specific heat of water is 4200 J kg^-1K^-1 , the specific heat of oil in the same limit is equal to 

As we know, work done by an electric heater

$\text{ i.e. }{m}_1{c}_1\mathrm{\Delta }t+m_2{c}_2\mathrm{\Delta }t=\text{ work done }$

So,    $m_1{c}_1\mathrm{\Delta }t+m_2{c}_2\mathrm{\Delta }t=P_1{t}_1$

where,  $m_1=0.5\mathrm{kg},\text{ specific heat, }{c}_1=4200{\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1}$

$\mathrm{\Delta }t=\mathrm{\Delta }{t}_1=\mathrm{\Delta }{t}_2=3\mathrm{K}$

$P_1=10\mathrm{W}$

$t_1=15\times 60=900\mathrm{s}$

c₂ = specific heat capacity of container.

$\text{ So, }0.5\times 4200\times (3-0)+m_2{c}_2\times (3-0)=10\times 15\times 60$

$\Rightarrow 2100\times 3+m_2{c}_2\times 3=9000$

$\Rightarrow m_2{c}_2=\frac{9000-6300}{3}=900$

Similarly, in the case of oil,  $m_1{c}_o\mathrm{\Delta }t+m_2{c}_2\mathrm{\Delta }t=P_2{t}_2$

$\text{ where, }{c}_o=\text{ specific heat capacity of oil, }{P}_2=10\mathrm{W}$

$2\times c_0\times 2+900\times 2=10\times 20\times 60$

$\Rightarrow 4c_0+1800=12000$

$\Rightarrow c_o=2550=2.55\times {10}^3{\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1}$