The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1 and T2 (T1 > T2). The rate of heat transfer, dQdt through the rod in a steady state is given by :
- dQdt = k(T1 - T2)LA
- dQdt = kLA(T1 - T2)
- dQdt = kA(T1 - T2)L
- dQdt = kL(T1 - T2)A
Answer
(3) dQdt = kA(T1 - T2)L
Solution
As per NCERT, in the steady state the rate of flow of heat through a rod is proportional to the temperature difference and the area of cross-section, and inversely proportional to the length:
dQdt = kA(T1 - T2)L
where k is the thermal conductivity of the material.
Answer: (3)