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, \(\dfrac{dQ}{dt}\) through the rod in a steady state is given by:
1
\(\dfrac{dQ}{dt}= \dfrac{k(T_1 - T_2)}{LA}\)
2
\(\dfrac{dQ}{dt}=k \ L \ A (T_1 - T_2)\)
3
\(\dfrac{dQ}{dt}=\dfrac{k \ A(T_1 - T_2)}{L}\)
4
\(\dfrac{dQ}{dt}= \dfrac{kL(T_1 - T_2)}{A}\)
5
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