The amount of heat transferred under steady-state across a slab of cross-section area 0.1 m² and thickness of 0.02 m with following assumptions

Material conductivity: 150 W/mK, Temperature gradient 20°C is

Concept:

The rate of heat transfer through a slab is given by:

\(\dot Q = \frac{{{\rm{\Delta T}}}}{{\sum \frac{{\rm{L}}}{{{\rm{K}} \times {\rm{A}}}}}}\)

where L = thickness of the wall, k = thermal conductivity, A = surface area of the wall

Calculation:

Given:

Temperature gradient = 20°C, K = 150 W/mK, A = 0.1 m2

Δ T = 20 °C

The rate of heat transfer through this wall is

\(\dot Q = \frac{{{\rm{\Delta T}}}}{{\sum \frac{{\rm{L}}}{{{\rm{K}} \times {\rm{A}}}}}}\)

\(\dot Q = \frac{{20}}{{\frac{{0.02}}{{150\times0.1}}}} = 15000~W = 15\;kW\)

Hence the required rate of heat transfer will be 15 kW.