Question

A body cools from ${80}^{o}C$ to ${64}^{o}C$ in $5min$ and same body cools from ${80}^{o}C$ to ${52}^{o}C$ in $10min$, what is the temperature of the surrounding ?

• A. ${24}^{o}C$
• B. ${28}^{o}C$
• C. ${22}^{o}C$
• D. ${25}^{o}C$

Answer 1

Answer: (A)

${24}^{o}C$

Given

Body cool from 80C to 64C in 5 min

Same Body cool from 80C to 52C in 10 min

Solution

Newton law of cooling

$\frac{dT}{dt}=k[T_t-T_s]$

where $T_t$ is average temperature between two temperature

And $T_s$ is temperature of surroundings

For Case 1

$\frac{80-64}{5}=k[\frac{80+64}{2}-T_s]$

Case 2

$\frac{80-52}{10}=k[\frac{80+52}{2}-T_s]$

Dividing both

$\frac{16}{14}=\frac{72-T_s}{66-T_s}$

$T_s=24C$

The correct option is A