Consider two rods of same length and different specific heats (S1, S2), conductivities (K1, K2) and area of cross-sections (A1, A2) and both having temperatures T1 and T2 at their ends. If rate of loss of heat due to conduction is equal, then
- K1A1 = K2A2
- K1A1S1 = K2A2S2
- K2A1 = K1A2
- K2A1S2 = K1A2S1
Answer
(1) K1A1 = K2A2
Solution
Rate of heat flow by conduction: Qt = KA(T1 - T2)L
L and (T1 - T2) are the same for both rods and the rates are equal
K1A1(T1 - T2)L = K2A2(T1 - T2)L
K1A1 = K2A2 (specific heat does not enter the steady-state rate)
Answer: (1)