The amount of heat needed to heat $200 \ g$ of ice at $-10^{\circ}C$ to convert it into water at $30^{\circ}C$ is:
Specific heat capacity of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$
Specific heat capacity of water $= 4186 \ J \ kg^{-1} \ K^{-1}$
Latent heat of fusion of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ (in $J$)

One $kg$ of water, at $20, ^oC$, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20, \Omega$. The rms voltage in the mains is $200, V$. Ignoring heat loss from the kettle, the time taken for water to evaporate fully is close to.......... $min$. [Specific heat of water $= 4200, J/kg, ^oC$, Latent heat of water $= 2260, kJ/kg$]

$A$ 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 $Al = 0.215 \, cal/g^o C$)

$A$ boy is able to chew $20 \, g$ of ice cream in $5 \, minutes$. Calculate the power of the boy in horse power $(HP)$. (Assume the latent heat of fusion of ice is $80 \, cal/g$ and $1 \, cal = 4.2 \, J$).

$A$ beaker of height $H$ is made up of a material whose coefficient of linear thermal expansion is $3\alpha$. It is filled up to the brim by a liquid whose coefficient of thermal expansion is $\alpha$. If now the beaker along with its contents is uniformly heated through a small temperature $T$,the level of liquid will reduce by (given $\alpha T << 1$):

$A$ $10.0 \,W$ electrical heater is used to heat a container filled with $0.5 \,kg$ of water. It is found that the temperature of the water and the container rose by $3 \,K$ in $15 \,min$. The container is then emptied,dried,and filled with $2 \,kg$ of an oil. It is now observed that the same heater raises the temperature of the container-oil system by $2 \,K$ in $20 \,min$. Assuming no other heat losses in any of the processes,the specific heat capacity of the oil is ................ $\times 10^3 \,J K^{-1} kg^{-1}$.