Lesson 7 - Radiation and Combined SAQ

** Calculate the heat loss due to radiation from the surface of a 100 hL copper kettle at 100°C to a room at 20°C. The surface area of the kettle is 27.5 m² and the emissivity of copper is 0.03.

Response:

The Stefan-Boltzmann law adapted for net rate of heat exchange between two bodies is:

Where

q is the heat transfer rate by radiation in Js^-1

A is the heat transfer area in m²

σ is the Stefan-Boltzmann constant = 5.672 × 10^-8 Wm^-2 K^-4

ε is the emissivity of the surface

T₁ and T₂ are the temperatures of the emitting surface and surrounding matter in K

Using this equation and the given values in the problem statement, we convert the temperatures to the appropriate units of Kelvin and then we can simply enter the values and calculate.

So, the kettle at 100°C is losing 561 J per second of heat to the atmosphere at 20°C. To prevent this heat loss and to protect personnel from the risk of burns from the hot vessel surface it would be sensible to fit the vessel with suitable insulation.