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Q.
If a metallic sphere gets cooled from $ 62^{\circ}C $ to $ 50^{\circ}C $ in $ 10 $ min and in the next $ 10 $ min gets cooled to $ 42^{\circ}C $ , then the temperature of the surroundings is
We know that rate of cooling
$\frac{T_{1}-T_{2}}{t}=K\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right]$
$\therefore \frac{62-50}{10}=K\left[\frac{62+50}{2}-T_{0}\right]$
$\Rightarrow \frac{12}{10}=K\left[56-T_{0}\right]$ ...(i)
and $\frac{50-42}{10}=K\left[\frac{50+42}{2}-T_{0}\right]$
$\Rightarrow \frac{8}{10}=K\left[46-T_{0}\right]$ ...(ii)
Solving Eqs. (i) and (ii),
we get $T_{0}= 26^{\circ} C$