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 $T_0=33.3^{\circ }C$