Question

If a metallic sphere gets cooled from $62^{\circ }C$ to $50^{\circ }C$ in $10$ min and in the next $10$ min gets cooled to $42^{\circ }C$ , then the temperature of the surroundings is

• A. $30^{\circ }C$
• B. $36^{\circ }C$
• C. $26^{\circ }C$
• D. $20^{\circ }C$

Answer 1

Answer: (C)

We know that rate of cooling
$\frac{T_1-T_2}{t}=K\left[\frac{T_1+T_2}{2}-T_0\right]$
$\therefore \frac{62-50}{10}=K\left[\frac{62+50}{2}-T_0\right]$
$\Rightarrow \frac{12}{10}=K\left[56-T_0\right]$ ...(i)
and $\frac{50-42}{10}=K\left[\frac{50+42}{2}-T_0\right]$
$\Rightarrow \frac{8}{10}=K\left[46-T_0\right]$ ...(ii)
Solving Eqs. (i) and (ii),
we get $T_0=26^{\circ }C$