Question

Consider two rods of same length and different specific heats (s₁ and s₂), conductivities K₁ and K₂ and areas of cross-section (A₁ and A₂) and both giving temperature T₁ and T₂ at their ends. If the rate of heat loss due to conduction is equal, then

• A. $K_{1}A_{1} = K_{2}A_{2}$
• B. $K_{2}A_{1} = K_{1}A_{2}$
• C. $\frac{K_{1}A_{1}}{s_{1}} = \frac{K_{2}A_{2}}{s_{2}}$
• D. $\frac{K_{2}A_{1}}{s_{2}} = \frac{K_{1}A_{2}}{s_{1}}$

Answer 1

Answer: (A)

$K_{1}A_{1} = K_{2}A_{2}$

Answer 2

According to problem, rate of heat loss in both rods are equal i.e. $\left( \frac{dQ}{dt} \right)_{1} = \left( \frac{dQ}{dt} \right)_{2}$

⇒ $\frac{K_{1}A_{1}\Delta\theta_{1}}{l_{1}} = \frac{K_{2}A_{2}\Delta\theta_{2}}{l_{2}}$

∴ $K_{1}A_{1} = K_{2}A_{2}$ [As $\Delta\theta_{1} = \Delta\theta_{2}$= (T₁ – T₂) and

$l_{1} = l_{2}$ given]