Question

Compare the rate of heat transfer by radiation for two objects, the first one has the temperature Ts1= 25 degree Celsius and the second is kept at temperature Ts2 = 40 degree Celsius. Suppose they are made of identical material (e1=e2) and have the same area participating in radiation (Ar1=Ar2). The Surrounding temperature Tr= 25 degree Celsius.

Answer 1

The rate of heat transfer of the second object is greater than the first object.

Step-by-step explanation:

`\varepsilon` = Emissivity of the object

`\sigma` = Stefan-Boltzmann constant =
`5.67* 10^(-8)\ \text{W/m}^2/\text{K}^4`

`T_1` = Temperature of surface 1 =
`25^(\circ)\text{C}+273.15=298.15\ \text{K}`

`T_2` = Temperature of surface 2 =
`40^(\circ)\text{C}+273.15=313.15\ \text{K}`

`T_0` = Surrounding temperature =
`25^(\circ)\text{C}+273.15=298.15\ \text{K}`

Rate of heat transfer is given by

`P_1=\varepsilon \sigma (T_1^4-T_0^4)\\\Rightarrow P_1=\varepsilon \sigma A_1(298.15^4-298.15^4)\\\Rightarrow P_1=0\ \text{W}`

`P_2=\varepsilon \sigma (T_2^4-T_0^4)\\\Rightarrow P_2=\varepsilon 5.67*10^(-8) A_2(313.15^4-298.15^4)\\\Rightarrow P_2=\varepsilon A_297.2`

`\varepsilon A_297.2>0`

So,
`P_2>P_1`

Hence, the rate of heat transfer of the second object is greater than the first object.