Question

A liquid takes $5$ minutes to cool from 80${}^o$C to 50${}^o$C. How much time will it take to cool from 60${}^o$C to 30${}^o$C? (Temperature of the surrounding= 20${}^o$C)

• A. 10 minutes
• B. 15 minutes
• C. 5 minutes
• D. 7.5 minutes

Answer 1

The correct option is A 10 minutes

From the fourier's law we know that $Q=-KA\frac{d\theta }{dt}$
On integrating it we find, $t=\frac{1}{K}\log \left(\frac{T_1-T_0}{T_2-T_0}\right)$
From the above relation, we will find the value of $K$ and will substitute it in second condition to find the time $t_2$
$5=\frac{1}{K}\log \left(\frac{80-20}{50-20}\right)$

So, $K=0.0602$

Now, $t_2=\frac{1}{0.0602}\log \left(\frac{60-20}{30-20}\right)$

So, $t_2=10.036=10$