consider the slab shown in Fig. 23-2 Suppose that ∆x = 24.9 cm, A = 1.80 m², and the material is copper. If T = - 12.0ºC, ∆T =136 Cº, and a steady state is reached, find (a) the temperature at a point in the rod 11.0 cm from the high-temperature and.

We know that temperature gradient is,
ΔT/Δx = 1643 K/m
(T_H – T)/Δx = 1643 K/m (Since, ΔT = T_H – T)
T_H – T = (1643 K/m) Δx
T = T_H - (1643 K/m) Δx
To find out T, substitute 670 K for T_H and 11.0 cm for Δx in the equation T = T_H - (1643 K/m) Δx,
T = T_H - (1643 K/m) Δx
= (670 K) – ((1643 K/m) (11.0 cm))
= (670 K) – ((1643 K/m) (11.0 cm))
= (670 K) – ((1643 K/m) (11.0 cm×10^-2 m/1 cm ))
= (670 K) – ((1643 K/m) (0.11 m)
= (670 K) – (180.73 K)
= 489.27 K
Rounding off to two significant figures, the temperature at a point in the rod 11.0 cm from the high temperature end will be 490 K.