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Question

Physics Question on Calorimetry

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0^{\circ}C and 90^{\circ}C respectively. The temperature of junction of the three rods will be

A

45C^{\circ}C

B

60C^{\circ}C

C

30C^{\circ}C

D

20C^{\circ}C

Answer

60C^{\circ}C

Explanation

Solution

Let θ\theta be the temperature of the junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow through AB + Rate of heat flow through CB = Rate of heat flow through BD 90θR+90θR=θ0R\therefore \, \, \, \, \, \frac{90^{\circ}-\theta}{R}+\frac{90 ^{\circ}-\theta}{R}=\frac{\theta -0}{R} Here, R = Thermal resistance 3θ=180orθ=60\therefore \, \, \, \, \, \, 3\theta = 180^{\circ} \, or \, \, \theta=60^{\circ} NOTE Rate of heat flow (H)=Temperaturedifference(TD)Thermalresistance(R \, \, \, \, \, (H) =\frac{Temperature \, difference \, (TD)}{ Thermal \, resistance \, (R} where, R=lKA \, \, \, R=\frac{l}{KA} K = Thermal conductivity of the rod. This is similar to the current flow through a resistance (R) where current (i) = Rate of flow of charge Potentialdifference(PD)Electricalresistance(R)\frac{Potential \, difference \, (PD)}{Electrical \, resistance \, (R)} Here, R =1σA=\frac{1}{\sigma A} where σ\sigma = Electrical conductivity