Gujarati
10-2.Transmission of Heat
hard

A liquid cools from $50^oC$ to $45^oC$ in 5 minutes and from $45 ^o C$ to $41.5 ^o C$ in the next $5$ minutes. The temperature of the surrounding is ...... $^oC$

A

$27$

B

$40.3$

C

$23.3 $

D

$33.3$

Solution

(d)$\frac{{50 – 45}}{5} = K\left( {\frac{{50 + 45}}{2} – {\theta _0}} \right)$ ….$.(i)$
$\frac{{45 – 41.5}}{5} = K\left( {\frac{{45 + 41.5}}{2} – {\theta _0}} \right)$…..$(ii)$
Solving equation $(i)$ and $(ii)$ we set ${\theta _0} = 33.3\;^\circ C$.

Standard 11
Physics

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