The rate of heat absorption in the rod would be : | TestHub

Passage / Comprehension

For heat flow through a cross-section, $\frac{d Q}{d t} = - K S (\frac{d T}{d x})$ , where K is thermal conductivity, S is cross-sectional area and $\frac{d T}{d x}$ is the temperature gradient. In the rod, $T (x) = 100 - 2 x^{2} - 4 x$ kelvin, with x in metres. Length $L = 2 m$ ; $K = 100 W/ (m K)$ ; $S = 0.1 m^{2}$ ; $M = 5 kg$ ; $α = 1 0^{- 4} / K$ . The source diagram marks A at $x = 0$ and B at $x = L$ .

The rate of heat absorption in the rod would be :

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Answer:

Solution:

(14 and 15)

$\frac{dQ}{dt} = - KS (\frac{dT}{dx})$

$\frac{dT}{dx} = - 4 x - 4$

$(\frac{dT}{dx})_{A} = - 4 x$

$(\frac{dT}{dx})_{B} = - 12 \frac{K}{m}$

Write q for the heat-flow rate at each end.

$q = \frac{dQ}{dt}$

The source uses the following absorption convention:

$q_{abs} = q_{A} - q_{B}$

$q_{abs} = K S [- 4 + 12]$

$q_{abs} = 100 \times 0.1 \times 8$

$q_{abs} = 80 Watt$

Physics - Heat & Thermo Question with Solution | TestHub

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