Question

A liquid cools from $50^{\circ} C$ to $45^{\circ} C$ in $5$ minutes and from $45^{\circ} C$ to $41.5^{\circ} C$ in the nest 5 minutes. The temperature of the surrounding is

• A. $27^{\circ}$
• B. $40.3^{\circ}$
• C. $23.3^{\circ}$
• D. $33.3^{\circ}$

Answer 1

Answer: (D)

First Case :
$\therefore \frac{50^{\circ}-45^{\circ}}{5}$
$=\left(\frac{50^{\circ}+45^{\circ}}{2}-T_{0}\right)$ ...(1)
$T_{0}=$ temperature of surrounding
Second Case :
$\frac{45^{\circ}-41.5^{\circ}}{5}$
$=K\left(\frac{45^{\circ}+41.5^{\circ}}{2}-T_{0}\right)$ ...(2)
Dividing equation (1) by equation (2)
$\frac{50^{\circ}-45^{\circ}}{45^{\circ}-41.5^{\circ}}=\frac{\left(\frac{50^{\circ}+45^{\circ}}{2}-T_{0}\right)}{\left(\frac{45^{\circ}+41.5^{\circ}}{2}-T_{0}\right)}$
$\frac{5}{3.5}=\frac{\left(47.50-T_{0}\right)}{\left(43.25-T_{0}\right)}$
$\frac{10}{7}=\frac{\left(47.50-T_{0}\right)}{\left(43.25-T_{0}\right)}$
On solving, we get $T_{0}=33.3^{\circ} C$