Answer:
[tex]T_{1}=94.9^{o}C[/tex]
Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:
[tex]T_{1}=T_{2}+\frac{q}{kS}[/tex]
Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium
[tex]S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\[/tex]
Substitute the given values
[tex]S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m[/tex]
The temperature of heater is then:
[tex]T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C[/tex]
The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
[tex]T_{1}=94.9^{o}C[/tex]
