Answer:
The emissivity of the surface of this object is 0.809.
Step-by-step explanation:
Given that,
Diameter = 15.0 cm
Temperature = 112°C
Power = 71.3 W
We need to calculate the area
Using formula of area
![A=4\pi r^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w4855iy8lys88d70lhy7su079lw41yc7vh.png)
Put the value into the formula
![A=4\pi((15.0*10^(-2))^2)/(4)](https://img.qammunity.org/2020/formulas/physics/college/fpmzxk4qz47kpas4m3yuazjtyfyyukl2jz.png)
![A=0.0707\ m^2](https://img.qammunity.org/2020/formulas/physics/college/r8kec4aeuhe4kjwg5fdfzpp9wyqcz1ja7k.png)
We need to calculate the emissivity of the surface of this object
Using formula of the emissivity
![P=\epsilon \sigma AT^4](https://img.qammunity.org/2020/formulas/physics/college/nevrlig6dov8o90g5n02t7y4in8ptuuyxb.png)
![\epsilon=(P)/(\sigma AT^4)](https://img.qammunity.org/2020/formulas/physics/college/o3a66lpijlpr2jkbc2e367thwen8q6w1fe.png)
Where, A = area
T = temperature
=Stefan Boltzmann constant
P = power
Put the value into the formula
![\epsilon=(71.3)/(5.670*10^(-8)*0.0707*(112+273)^4)](https://img.qammunity.org/2020/formulas/physics/college/hp5et4syzw2rgjwpplnuk02rmcncimk2an.png)
![\epsilon=0.809](https://img.qammunity.org/2020/formulas/physics/college/d9u7m1mfgq4mfd6tcst7bcflg6ykqcyc0t.png)
Hence, The emissivity of the surface of this object is 0.809.