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Q.
A body takes $10$ minutes to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of surroundings is constant at $25^{\circ}C$. Then, the temperature of the body after next $10$ minutes will be approximately :
Time taken to cool from $60^{\circ} C$ to $50^{\circ} C =10$ minutes
Temperature of surroundings $=25^{\circ} C$ Temperature of body in next 10 minutes $=T$
Therefore, $\frac{60-50}{10 \min }=k_{ B }\left[\frac{60+50}{2}-25\right] \Rightarrow k_{ B } 30=1\,\,\,\,\,\,\,(1)$
and $\frac{60-T}{20 \min }=k_{ B }\left[\frac{60+T}{2}-25\right]=k_{ B }\left[\frac{60+T-50}{2}\right]\,\,\,\,\,\,\,(2)$
Taking ratio of Eqs. (1) and (2), we get
$\frac{20}{60-T}=\frac{30 k_{ B }}{k_{ B }\left(\frac{10+T}{2}\right)} $
$\Rightarrow \frac{20}{60-T}=\frac{30}{5+T / 2}$
$\Rightarrow 20\left(5+\frac{T}{2}\right)=30(60-T) $
$\Rightarrow 100+T\, 10=1800-30 \,T $
$\Rightarrow 1800-100=30\, T+10 \,T$
$\Rightarrow 1700=40 \,T $
$\Rightarrow T=\frac{1700}{40}=42.5^{\circ} C -43^{\circ} C$