Question

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 :

• A. $47^{\circ }C$
• B. $41^{\circ }C$
• C. $45^{\circ }C$
• D. $43^{\circ }C$

Answer 1

Answer: (D)

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
$\frac{20}{60-T}=\frac{30k_B}{k_B\left(\frac{10+T}{2}\right)}$
$\Rightarrow \frac{20}{60-T}=\frac{30}{5+T/2}$
$\Rightarrow 20\left(5+\frac{T}{2}\right)=30(60-T)$
$\Rightarrow 100+T10=1800-30T$
$\Rightarrow 1800-100=30T+10T$
$\Rightarrow 1700=40T$
$\Rightarrow T=\frac{1700}{40}=42.5^{\circ }C-43^{\circ }C$