Answer to Question #140238 in Molecular Physics | Thermodynamics for Sridhar

Question

A solid cylinder of radius r(1)=2.5cm length L(1)=5.0cm and temperature 40°C is suspended in an environment of temperature 60°C .The thermal radiation transfer rate for cylinder is 1.0 Watt.If the cylinder is stretched until its radius becomes r(2)=0.50 cm the thermal radiation transfer rate is changed to
Ans 3.35W

Answer 1

$a) P = e\times A\times (T2^4 - T1^4) = e\times (2\times pi\times r1^2+2\times pi\times r1\times L1) \times (T2^4 - T1^4)=125.309 W$

B)- as volume is constant

$r1^2\times L1 = r2^2 \times L2$

$L2 = 1.25m$

$P2 = e\times (2\times pi\times r2\times L2) \times (T2^4 - T1^4) = 459.46$$W$

$\frac{P2}{P1} = \frac{459.46}{125.309}$ = 3.67