Answer to Question #141281 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 1.0Watt. if the cylinder is stretched until its radius becomes r(2)=0.50cm the thermal radiation transfer rate is changed to :
Ans 3.35w

Answer 1

r₁ = 0.025 m

T = 40 + 273 = 313 K

L₁ = 0.05 m

T_env = 60 + 273 = 333 K

P₁ = 1.0 Watt

r₂ = 0.005 m

P₁ = σεA₁(T_env⁴ – T⁴)

P₂ = σεA₂(T_env⁴ – T⁴)

$σ = 5.6704 \times 10^{-8} \; \frac{W}{m^2K^4}$ Stefan Boltzmann constant

$A_1 = πr_1^2 + πr_1^2 +(2 πr_1)L_1$

$A_1 = 3.14 \times 0.025^2 + 3.14 \times 0.025^2 +(2 \times 3.14 \times 0.025) \times 0.05 = 0.011780972 \;m^2$

$ε = \frac{P_1}{σA_1(T_{env}^4 – T^4)}$

$ε = \frac{1}{(5.6704 \times 10^{-8})(0.011780972)(333^4 – 313^4)} = 0.555$

The stretched cylinder and the former have the same volume:

V₁ =V₂

πr₁²L₁ = πr₂²L₂

$L_2 = \frac{πr_1^2L_1}{πr_2^2}$

$L_2 = \frac{0.025^2 \times 0.05}{0.005^2} = 1.25 \;m$

$A_2 = πr_2^2 + πr_2^2 +(2 πr_2)L_2$

$A_2 = 3.14 \times (0.005)^2 + 3.14 \times (0.005)^2 + (2\times 3.14 \times 0.005) \times 1.25 = 0.039407 \;m^2$

P₂ = σεA₂(T_env⁴ – T⁴)

$P_2 = (5.6704 \times 10^{-8})(0.555)(0.039407)(333^4 – 313^4) = 3.346\; W$