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

Answer:

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

Explanation:

$\varepsilon$ = Emissivity of the object

$\sigma$ = Stefan-Boltzmann constant = $5.67\times 10^{-8}\ \text{W/m}^2/\text{K}^4$

$T_1$ = 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

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

$P_2=\varepsilon \sigma (T_2^4-T_0^4)\\\Rightarrow P_2=\varepsilon 5.67\times10^{-8} A_2(313.15^4-298.15^4)\\\Rightarrow P_2=\varepsilon A_297.2$

$\varepsilon A_297.2>0$

So, $P_2>P_1$

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