JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 20)
A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 mthick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -
200 Wm$$-$$2
65 Wm$$-$$2
120 Wm$$-$$2
90 Wm$$-$$2
Explanation
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$$\left( {{{dQ} \over {dt}}} \right) = {{kA\Delta T} \over \ell }$$
$$ \Rightarrow $$ $${1 \over A}\left( {{{dQ} \over {dt}}} \right) = {{\left( {0.1} \right)\left( {900} \right)} \over 1} = 90W/{m^2}$$
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