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$)

$A$ 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. $A$ cubical icebox of side $30 \,cm$ has a thickness of $5.0 \,cm$. If $4.0 \,kg$ of ice is put in the box,estimate the amount of ice (in $kg$) remaining after $6 \,h$. The outside temperature is $45 \,^{\circ}C$,and the coefficient of thermal conductivity of thermacole is $0.01 \,J \,s^{-1} \,m^{-1} \,K^{-1}$. Heat of fusion of water $= 335 \times 10^{3} \,J \,kg^{-1}$.

The temperature difference across two cylindrical rods $A$ and $B$ of the same material and same mass are $40^{\circ} C$ and $60^{\circ} C$ respectively. In steady state,if the rates of flow of heat through the rods $A$ and $B$ are in the ratio $3: 8$,the ratio of the lengths of the rods $A$ and $B$ is (in $: 3$)

Two identical rods $AC$ and $CB$ made of two different metals having thermal conductivities in the ratio $2:3$ are kept in contact with each other at the end $C$. $A$ is at $100^{\circ}C$ and $B$ is at $25^{\circ}C$. Then the junction $C$ is at: (in $^{\circ}C$)

$A$ thermocol box has a total wall area (including the lid) of $1.0 \,m^2$ and wall thickness of $3 \,cm$. It is filled with ice at $0^{\circ} C$. If the average temperature outside the box is $30^{\circ} C$ throughout the day, the amount of ice that melts in one day is (Given: $K_{\text{thermocol}} = 0.03 \,W/mK$, $L_{\text{fusion(ice)}} = 3.00 \times 10^5 \,J/kg$) (in $\,kg$)

Three rods of the same dimensions have thermal conductivities $3K, 2K$, and $K$. They are arranged as shown in the figure. The temperature of the junction in the steady state is: