Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same, and if the rate of flow of heat becomes $H_2$, then $H_2 =$

The temperature of the water of a pond is $0^{\circ} C$ while that of the surrounding atmosphere is $-20^{\circ} C$. If the density of ice is $\rho$, the coefficient of thermal conductivity is $k$, and the latent heat of melting is $L$, then the thickness $Z$ of the ice layer formed increases as a function of time $t$ as:

The rate of heat conduction through a cylindrical rod is $Q_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod are doubled while keeping the temperatures the same,the new rate of heat conduction is $Q_2$. Then:

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2}$ $(T_{1} > T_{2})$. The rate of heat transfer,$\frac{dQ}{dt}$ through the rod in a steady state is given by:

Rate of flow of heat through a cylindrical rod is $H_1$. The temperatures at the ends of the rod are $T_1$ and $T_2$. If all the dimensions of the rod become double and the temperature difference remains the same,the rate of flow of heat becomes $H_2$. Then:

If $4 \, kg$ of ice is inside a closed cubical thermocol box of side length $20 \, cm$ and wall thickness $4 \, cm$, then the mass of the ice remaining after $10 \, hours$ is nearly. (Given: Outside temperature $= 50^{\circ}C$, Thermal conductivity of thermocol $K = 0.01 \, Js^{-1}m^{-1\circ}C^{-1}$, Latent heat of fusion of ice $L = 335 \times 10^3 \, Jkg^{-1}$) (in $ \, kg$)