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$)

$A$ copper pipe of length $10 \,m$ carries steam at temperature $110^{\circ} C$. The outer surface of the pipe is maintained at a temperature $10^{\circ} C$. The inner and outer radii of the pipe are $2 \,cm$ and $4 \,cm$,respectively. The thermal conductivity of copper is $0.38 \,kW / m /^{\circ} C$. In the steady state,the rate at which heat flows radially outward through the pipe is closest to ............. $\,kW$.

$A$ slab of stone of area $0.36\, m^2$ and thickness $0.1\, m$ is exposed on the lower surface to steam at $100^{\circ} C$. $A$ block of ice at $0^{\circ} C$ rests on the upper surface of the slab. In one hour $4.8\, kg$ of ice is melted. The thermal conductivity of the slab is .......... $J/m/s/^{\circ} C$ (Given latent heat of fusion of ice $= 3.36 \times 10^5\, J/kg$)

$A$ rectangular ice box of total surface area of $1000 \,cm^2$ initially contains $1.5 \,kg$ of ice at $0^{\circ}C$. If the thickness of the walls of the box is $2 \,mm$ and the temperature outside the box is $42^{\circ}C$, then the mass of the ice remaining in the box after $160 \,minutes$ is (Thermal conductivity of the material of the box $= 10^{-2} \,W m^{-1} K^{-1}$ and latent heat of the fusion of ice $= 336 \times 10^3 \,J kg^{-1}$) (in $kg$)

If the radius and length of a copper rod are both doubled,the rate of flow of heat along the rod increases ....... times.

Four identical conducting rods are joined to form a square $ABCD$. The temperatures of junctions $A, B,$ and $C$ are maintained at $100^{\circ}C, 40^{\circ}C,$ and $0^{\circ}C$ respectively. Choose the $\text{INCORRECT}$ statement for this arrangement in a steady state. (The value of $\frac{KA}{L} = 1 \text{ J}/(s \cdot ^{\circ}C)$).