Question

A room is at a constant temperature of 300 K. A hotplate in the room is at a temperature of 400 K and loses energy by radiation at a rate of P. What is the rate of loss of energy from the hotplate when its temperature is 500 K?

Answer 1

(D) `P' = (\frac{5^4-3^4}{4^4-3^4})P`

Explanation:

A body with a non-zero temperature simultaneously emits and absorbs power. So the Net Thermal Power Loss is the difference between the total thermal power radiated by the object and the total thermal power power it absorbs from the surroundings.

`P_{Net} = P_{rad} - P_{abs}`,
`P_{Net} = \sigma A T^4 - \sigma A T_a^4 = \sigma A (T^4-T_a^4)`

where,
`T` - Temperature of the body (in Kelvins);
`T_a` - Temperature of the surroundings (in Kelvins),
`A` - Surface Area of the radiating object (in `m^2`),
`\sigma` - Stefan-Boltzmann Constant.

`P = \sigma A (400^4-300^4);`
`P' = \sigma A (500^4-300^4);`

`(P')/P =\frac{\cancel{\sigma A} (500^4-300^4)}{\cancel{\sigma A}(400^4-300^4)} =\frac{5^4-3^4}{4^4-3^4}`

`P' = (\frac{5^4-3^4}{4^4-3^4})P`