A cylindrical rod with a length of 50 cm and a cross-sectional area of 1 cm^2 is positioned between a large ice chamber at 0^ ^ ^ C and an evacuated chamber kept at 27^ ^ ^ C , as shown in the figure. Very small

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1.8 W m^ -1 K^ -1

In steady state, the net rate of heat absorbed by the blackened end via radiation equals the rate of heat conducted through the rod.

The net rate of heat absorbed by radiation is given by the Stefan-Boltzmann law:

P_{\text{rad}} = \sigma A (T_0^4 - T_1^4)

where

T_0 = 27^{}^{{}^{\circ}}\text{C} = 300 \text{ K}

is the temperature of the evacuated chamber, and

T_1 = 17^{}^{{}^{\circ}}\text{C} = 290 \text{ K}

is the temperature of the blackened end.

The rate of heat conduction through the rod is:

P_{\text{cond}} = \dfrac{k A (T_1 - T_2)}{L}

where

T_2 = 0^{}^{{}^{\circ}}\text{C} = 273 \text{ K}

is the temperature of the ice chamber, and

L = 50 \text{ cm} = 0.5 \text{ m}

.

Equating

P_{\text{rad}}

and

P_{\text{cond}}

:

\sigma A (T_0^4 - T_1^4) = \dfrac{k A (T_1 - T_2)}{L}

\Rightarrow k = \dfrac{\sigma L (T_0^4 - T_1^4)}{T_1 - T_2}

Substituting the given values:

T_0^4 - T_1^4 = (300)^4 - (290)^4 = 81 \times 10^8 - 70.7281 \times 10^8 = 10.2719 \times 10^8 \text{ K}^4

k = \dfrac{6 \times 10^{-8} \times 0.5 \times 10.2719 \times 10^8}{290 - 273}

k = \dfrac{3 \times 10.2719}{17} = \dfrac{30.8157}{17} \approx 1.8 \text{ W m}^{-1} \text{ K}^{-1}

Answer:

1.8 \text{ W m}^{-1} \text{ K}^{-1}

A cylindrical rod with a length of 50 cm and a cross-sectional area...

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