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10-2.Transmission of Heat
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A body of length 1m having cross sectional area $0.75\;m^2$ has heat flow through it at the rate of $ 6000\; Joule/sec$ . Then find the temperature difference if $K = 200\;J{m^{ - 1}}{K^{ - 1}}$ ...... $^oC$
A
$20$
B
$40$
C
$80$
D
$100$
Solution
(b) $\frac{Q}{t} = \frac{{KA\Delta \theta }}{l}$ ==> $6000 = \frac{{200 \times 0.75 \times \Delta \theta }}{1}$
$\therefore$ $\Delta \theta = \frac{{6000 \times 1}}{{200 \times 0.75}} = 40^\circ C$
Standard 11
Physics
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