An aluminium container of mass $100 \, g$ contains $200 \, g$ of ice at $-20^{\circ} C$. Heat is added to the system at the rate of $100 \, cal/s$. The temperature of the system after $4$ minutes will be ....... $^{\circ} C$ (specific heat of ice $= 0.5 \, cal/g^{\circ} C$,latent heat of fusion $L = 80 \, cal/g$,specific heat of $Al = 0.2 \, cal/g^{\circ} C$,specific heat of water $= 1 \, cal/g^{\circ} C$)

The freezing point of a liquid decreases when pressure is increased,if the liquid:

The heat energy required to convert $10 \,kg$ of ice at $-10^{\circ} C$ into water at $0^{\circ} C$ is (specific heat capacity of ice $=0.5 \,cal \,g^{-1} \,^{\circ}C^{-1}$ and latent heat of fusion of ice $=80 \,cal \,g^{-1}$)

$A$ metal ball immersed in alcohol weighs $W_1$ at $0^{\circ}C$ and $W_2$ at $59^{\circ}C$. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol,it can be shown that

$A$ copper ring has a diameter of exactly $25 \, mm$ at its temperature of $0^o C$. An aluminium sphere has a diameter of exactly $25.05 \, mm$ at its temperature of $100^o C$. The sphere is placed on top of the ring and the two are allowed to come to thermal equilibrium,with no heat being lost to the surroundings. The sphere just passes through the ring at the equilibrium temperature. The ratio of the mass of the sphere to the ring is: (Given: $\alpha_{Cu} = 17 \times 10^{-6} /^o C$,$\alpha_{Al} = 2.3 \times 10^{-5} /^o C$,specific heat of $Cu = 0.0923 \, cal/g^o C$,and specific heat of $Al = 0.215 \, cal/g^o C$)

$A$ clock pendulum having a coefficient of linear expansion $\alpha = 9 \times 10^{-7} /^{\circ}C$ has a period of $0.5 \ s$ at $20^{\circ}C$. If the clock is used in a climate where the temperature is $30^{\circ}C$,how much time does the clock lose in each oscillation? (Assume $g$ is constant)