(B) The rate of heat flow through the plates in series is given by:$\frac{Q}{t} = \frac{K_1 A (\theta_1 - \theta)}{L_1} = \frac{K_2 A (\theta - \theta

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(B) The rate of heat flow through the plates in series is given by:$\frac{Q}{t} = \frac{K_1 A (\theta_1 - \theta)}{L_1} = \frac{K_2 A (\theta - \theta_2)}{L_2}$Since the plates are of the same material,$K_1 = K_2 = K$. Also,the area $A$ is the same.Given: $\theta_1 = -20^{\circ}C$,$\theta_2 = 20^{\circ}C$,$L_1 = 2 \, cm$,$L_2 = 5 \, cm$.Substituting these values into the equation:$\frac{K A (-20 - \theta)}{2} = \frac{K A (\theta - 20)}{5}$$\frac{-20 - \theta}{2} = \frac{\theta - 20}{5}$$5(-20 - \theta) = 2(\theta - 20)$$-100 - 5\theta = 2\theta - 40$$7\theta = -60$$\theta = -\frac{60}{7} \approx -8.6^{\circ}C$Thus,the temperature of the contact surface is $-8.6^{\circ}C$.

Class 11 Physics 10-2.Heat Transfer Thermal Resistance and it's Combination

Medium

Two plates of equal area are placed in contact with each other. Their thicknesses are $2.0 \, cm$ and $5.0 \, cm$. The temperature of the outer surface of the first plate is $-20^{\circ}C$ and the temperature of the outer surface of the second plate is $20^{\circ}C$. If the plates are made of the same material,find the temperature of the contact surface in $^{\circ}C$.

A
$0$
B
$-8.6$
C
$20$
D
$25$

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10-2.Heat Transfer

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Thermal Resistance and it's Combination

Two rods of same material have same length and area. The heat $\Delta Q$ flows through them for $12 \, \text{minutes}$ when they are joined in series. If now both the rods are joined in parallel,then the same amount of heat $\Delta Q$ will flow in ........ $\text{minutes}$.

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Two rods of the same length and material transfer a given amount of heat in $12 \text{ s}$ when they are joined end-to-end. If they are joined lengthwise, in how many seconds will they transfer the same amount of heat under the same conditions (in $\text{ s}$)?

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Twelve identical conducting rods form the edges of a uniform cube of side length $l$. In the steady state,the junctions $B$ and $H$ are maintained at $100^{\circ}C$ and $0^{\circ}C$ respectively. Find the temperature of the junction $A$ in $^{\circ}C$.

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$A$ composite slab consists of two materials having coefficients of thermal conductivity $K$ and $2K$,thickness $x$ and $4x$ respectively. The temperatures of two outer surfaces of the composite slab are $T_2$ and $T_1$ respectively $(T_2 > T_1)$. The rate of heat transfer through the slab in a steady state is $\left[\frac{A(T_2 - T_1)K}{x}\right] f$,where $f$ is equal to:

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Two rods of same length, radius and material transfer a given amount of heat in '$t$' seconds when they are joined as shown in fig $(1)$. But when they are joined as shown in fig $(2)$, then they will transfer the same amount of heat under the same conditions in time: (in $t$)

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Two rods of same material have same length and area. The heat $\Delta Q$ flows through them for $12 \, \text{minutes}$ when they are joined in series. If now both the rods are joined in parallel,then the same amount of heat $\Delta Q$ will flow in ........ $\text{minutes}$.

Two rods of the same length and material transfer a given amount of heat in $12 \text{ s}$ when they are joined end-to-end. If they are joined lengthwise, in how many seconds will they transfer the same amount of heat under the same conditions (in $\text{ s}$)?

Twelve identical conducting rods form the edges of a uniform cube of side length $l$. In the steady state,the junctions $B$ and $H$ are maintained at $100^{\circ}C$ and $0^{\circ}C$ respectively. Find the temperature of the junction $A$ in $^{\circ}C$.

Two rods of same length, radius and material transfer a given amount of heat in '$t$' seconds when they are joined as shown in fig $(1)$. But when they are joined as shown in fig $(2)$, then they will transfer the same amount of heat under the same conditions in time: (in $t$)

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