$A$ rod of length $10 \, cm$ has a cross-sectional area of $100 \, cm^2$ and a thermal conductivity of $400 \, W/m^\circ C$. If the heat flow through the rod is $4000 \, J/s$,find the temperature difference between the two ends in $^\circ C$.

$A$ conducting rod of length $1 \,m$ has an area of cross-section $10^{-3} \,m^2$. One end is immersed in boiling water $(100^{\circ} C)$ and the other end in ice $(0^{\circ} C)$. If the coefficient of thermal conductivity of the rod is $96 \,cal/(s \cdot m \cdot ^{\circ}C)$ and the latent heat of fusion for ice is $8 \times 10^4 \,cal/kg$,then the amount of ice that will melt in one minute is:

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 $: 3$)

The two opposite faces of a cubical piece of iron (thermal conductivity $= 0.2 \text{ CGS units}$) are at $100^{\circ}C$ and $0^{\circ}C$ in ice. If the area of a surface is $4 \text{ cm}^2$,then the mass of ice melted in $10 \text{ minutes}$ will be ...... $\text{gm}$.

$A$ cylindrical rod has temperatures $T_1$ and $T_2$ at its ends. The rate of flow of heat is $Q_1 \text{ cal s}^{-1}$. If the length and radius of the rod are doubled while keeping the temperatures constant,then the new rate of flow of heat $Q_2$ will be:

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: