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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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