Answer:
The rate of heat transfer of the second object is greater than the first object.
Explanation:
[tex]\varepsilon[/tex] = Emissivity of the object
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ \text{W/m}^2/\text{K}^4[/tex]
[tex]T_1[/tex] = Temperature of surface 1 = [tex]25^{\circ}\text{C}+273.15=298.15\ \text{K}[/tex]
[tex]T_2[/tex] = Temperature of surface 2 = [tex]40^{\circ}\text{C}+273.15=313.15\ \text{K}[/tex]
[tex]T_0[/tex] = Surrounding temperature = [tex]25^{\circ}\text{C}+273.15=298.15\ \text{K}[/tex]
Rate of heat transfer is given by
[tex]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}[/tex]
[tex]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[/tex]
[tex]\varepsilon A_297.2>0[/tex]
So, [tex]P_2>P_1[/tex]
Hence, the rate of heat transfer of the second object is greater than the first object.
