$A$ copper rod of length $10 \ cm$ and cross-sectional area $100 \ cm^2$ is to conduct $4000 \ J/s$ of heat. The thermal conductivity of copper is $400 \ W/m \cdot ^\circ C$. The temperature difference between the two ends of the rod must be maintained at ............. $^\circ C$.

The area of the glass of a window of a room is $10\;m^2$ and thickness is $2\;mm$. The outer and inner temperatures are $40^{\circ}C$ and $20^{\circ}C$ respectively. The thermal conductivity of glass in the $MKS$ system is $0.2\;W/(m\cdot K)$. The heat flowing into the room per second will be:

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:

Two rods,one made of copper and the other made of steel,of the same length and same cross-sectional area,are joined together. The thermal conductivities of copper and steel are $385 \, W \, m^{-1} \, K^{-1}$ and $50 \, W \, m^{-1} \, K^{-1}$ respectively. The free ends of copper and steel are held at $100^{\circ} \, C$ and $0^{\circ} \, C$ respectively. The temperature at the junction is nearly $.......^{\circ} \, C$.

Dimensional formula for thermal conductivity is (here $K$ denotes the temperature)

$A$ slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio $1 : 4$. If the free face of brass is at $100^\circ C$ and that of copper at $0^\circ C$,the temperature of the interface is ........ $^\circ C$.