10 kg of ice at $-10^{\circ} \mathrm{C}$ is added to 100 kg of water to lower its temperature from 25 ${ }^{\circ} \mathrm{C}$. Consider no heat exchange to surroundings. The decrement to the temperature of water is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.

(specific heat of ice $=2100 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, specific heat of water $=4200 \mathrm{~J} / \mathrm{Kg} .{ }^{\circ} \mathrm{C}$, latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} / \mathrm{Kg}$ )

In the following $p-V$ diagram the equation of state along the curved path is given by $(V-2)^2=4 a p$ where $a$ is a constant. The total work done in the closed path is

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha$ Joule ( $P_1=21.7 \mathrm{~Pa}$ and $\left.P_2=30 \mathrm{~Pa}, \mathrm{C}_v=21 \mathrm{~J} / \mathrm{K} . \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$ .

Density of water at $4^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$ are $1000 \mathrm{~kg} / \mathrm{m}^3$ and $998 \mathrm{~kg} / \mathrm{m}^3$ respectively. The increase in internal energy of 4 kg of water when it is heated from $4^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$ is $\_\_\_\_$ J.

(specific heat capacity of water $=4.2 \mathrm{~J} / \mathrm{kg}$. and 1 atmospheric pressure $=10^5 \mathrm{~Pa}$ )