Question

One end of an iron poker is placed in a fire where the temperature is 651 °C, and the other end is kept at a temperature of 24.0 °C. The poker is 0.880 m long and has a radius of 4.40 × 10^-3 m. Ignoring the heat lost along the length of the poker, find the amount of heat conducted from one end of the poker to the other in 7.00 s. Take the thermal conductivity of iron to be 79 J/(s m C^o).

Answer 1

Answer:

Given, T1 = 651 0C, T2 = 24.0 0C so the temperature difference is T = (651 - 24) 0C = 627 0C.

Length of the iron poker L = 0.880 m, radius r = 4.40 x 10-3 m, time interval for heat conduction between the two ends is t = 7.00 s and thermal conductivity of iron k = 79 J/s.m 0C.

The amount of heat Q conducted (or heat loss) between two ends of the iron poker per unit time interval t per unit area A is directly proportional to the temperature gradient T/x.

Therefore, Q/t.A = -k T/x, where k is thermal conductivity of the material and "-" sign indicates the heat loss.

Here, x is equal to the length of the iron poker L and area of the poker is A = 2rL = 2(4.40 x 10-3 m) (0.880 m) = 2.43 x 10-2 m2.

Then, Q = kT A t / L = (79 J/s.m 0C) (627 0C) (2.43 x 10-2 m2) (7.00 s) / (0.880 m) = 9.58 x 103 J.