Samyak Jain
Last Activity: 5 Years ago
In steady state, heat current (i.e. heat per unit time) flowing throuh any cross section of conductur is same.
We know that dQ/dt = K A dT/dx ...(1), where dQ/dt is heat current, K = thermal coefficient of substance, A is area of cross section of length dx and dT/dx = temprature gradient.
Let point at T
1 = 20
C be A, that at T
2 = 0
C be B and that at 20cm from A be C with temperture T. AB = 60 cm.
CB = (60 – 20)cm = 40 cm.
In steady state, we can replace dx by length and dt by
T in (1) of conductor if area of cross-section is uniform.
dQ/dt across AC = dQ/dt across CB.
i.e. KA(T
1 – T)/20 cm = KA(T – T
2)/40 cm
(T – T
1)/20 + (T – T
2)/40 = 0. (by Kirchoff’s junction law)
2 T – 2 T
1 + T – T
2 = 0 or T = (2T
1 + T
2) / 3 = (2.20
C + 0
C)/3
T = (40 / 3) C .
