SATHEE: Heat and Thermodynamics 3 Question 31

Heat and Thermodynamics 3 Question 31

31. Two metal cubes $A$ and $B$ of same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of $A$ and $B$ are $300 W / m^{\circ} C$ and $200 W / m^{\circ} C$ , respectively. After steady state is reached the temperature $T$ of the interface will be

(1996, 2M)

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Solution:

Thermal resistance, $R = \frac{l}{K A}$ .

i.e. $R \propto \frac{1}{K}, l$ and $A$ being the same for both the blocks

$∴ \frac{R_{A}}{R_{B}} = \frac{K_{B}}{K_{A}} = \frac{200}{300} = \frac{2}{3}$

Rate of flow of heat (thermal current) along both the cubes will be equal (in series). Therefore, temperature difference across both the cubes will be in the ratio of their thermal resistances. Hence,

$\begin{aligned} \frac{(T D)_{A}}{(T D)_{B}} = \frac{R_{A}}{R_{B}} = \frac{2}{3} or \frac{100^{\circ} C - T}{T - 0^{\circ} C} = \frac{2}{3} \\ \Rightarrow 300^{\circ} C - 3 T = 2 T \Rightarrow T = 60^{\circ} C \end{aligned}$

Heat and Thermodynamics 3 Question 31

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Heat and Thermodynamics 3 Question 31

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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