Question

A food warmer made of thermo-plastic material is at 40°C and the surrounding environment is at 20°C. Calculate the rate of heat transfer per unit area of the surface,provided the surface is 20mm thick and the thermal conductivity of the material is 29W/m​

Answer 1

Answer: 870 W/m²

Step-by-step explanation:

Using Fourier's Law of Heat Conduction, the rate of heat transfer per unit area (q) can be calculated as:

q = k × (T1 - T2) / L

where k is the thermal conductivity of the material, T1 is the temperature of the warmer, T2 is the temperature of the surrounding environment, and L is the thickness of the material.

Plugging in the given values, we get:

q = 29 W/m·K × (40°C - 20°C) / (20 mm / 1000)

q = 870 W/m²

Therefore, the rate of heat transfer per unit area of the surface is 870 W/m².