Question

Consider a slab of face area A and thickness L. Suppose that L = 33 cm, A = 55 cm2, and the material is copper. If the faces of the slab are maintained at temperatures TH = 129°C and TC = 21°C, and a steady state is reached, find the conduction rate through the slab. The thermal conductivity of copper is 401 W/m·K.

Answer 1

721.8 Joule per second

Step-by-step explanation:

L = 33 cm = 0.33 m

A = 55 cm^2 = 0.0055 m^2

Th = 129°C

Tc = 21°C

k = 401 W/mK

The rate of flow of heat is given by the formula

`H =(K A \left ( T_(h)-T_(c) \right ))/(L)`

`H =\frac {401  * 0.0055 * \left (129-21\right )}{0.33}`

H = 721.8 Joule per second

Thus, the rate of flow of heat is given by 721.8 Joule per second.