(A) The rate of heat flow through a cylindrical shell is given by $H = \frac{Q}{t} = \frac{KA(T_1 - T_2)}{L}$.Here,the surface area $A = 2\pi r l = 2

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(A) The rate of heat flow through a cylindrical shell is given by $H = \frac{Q}{t} = \frac{KA(T_1 - T_2)}{L}$.Here,the surface area $A = 2\pi r l = 2 \times \pi \times 0.1 \times 2 = 0.4\pi \, m^2$.Given: $K = 390 \, W m^{-1} K^{-1}$,$T_1 = 373 \, K$,$T_2 = 273 \, K$,$L = 5 \, mm = 0.005 \, m$,and $t = 1 \, s$.Substituting the values:$Q = \frac{390 \times 0.4\pi \times (373 - 273) \times 1}{0.005}$$Q = \frac{390 \times 0.4 \times 3.14159 \times 100}{0.005}$$Q = \frac{156 \times 3.14159 \times 100}{0.005} \approx 98 \times 10^5 \, J$.

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

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Steam at $373 K$ is passed through a tube of radius $10 cm$ and length $2 m$. The thickness of the tube is $5 mm$ and the thermal conductivity of its material is $390 W m^{-1} K^{-1}$. Calculate the heat lost per second. The outside temperature is $0^{\circ}C$.

A
$98 \times 10^{5} J$
B
$85 \times 10^{5} J$
C
$95 \times 10^{5} J$
D
$80 \times 10^{5} J$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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:

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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:

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Two plates of equal area are placed in series. The ratios of their thicknesses and thermal conductivities are both $2:3$. The temperature of the outer surface of one plate is $100 ^\circ C$ and that of the other is $0 ^\circ C$. The temperature of the common surface is ....... $^\circ C$.

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Copper and iron rods of identical dimensions are coated with wax. The ratio of thermal conductivity of copper to iron is $10:9$. If one end of each rod is placed in hot water,the wax melts. Find the ratio of the lengths of the rods for which the wax melts at the same rate.

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Steam at $373 K$ is passed through a tube of radius $10 cm$ and length $2 m$. The thickness of the tube is $5 mm$ and the thermal conductivity of its material is $390 W m^{-1} K^{-1}$. Calculate the heat lost per second. The outside temperature is $0^{\circ}C$.

Similar Questions

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:

Two plates of equal area are placed in series. The ratios of their thicknesses and thermal conductivities are both $2:3$. The temperature of the outer surface of one plate is $100 ^\circ C$ and that of the other is $0 ^\circ C$. The temperature of the common surface is ....... $^\circ C$.

Copper and iron rods of identical dimensions are coated with wax. The ratio of thermal conductivity of copper to iron is $10:9$. If one end of each rod is placed in hot water,the wax melts. Find the ratio of the lengths of the rods for which the wax melts at the same rate.

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