Question

A body cools down from ${50}^0$C to ${45}^0$C in 5 minutes and to ${40}^0$C in another 8 minutes. Find the temperature of the surrounding.

Answer 1

${50}^{0}C...{.45}^{0}C...{.4}^0$ C
Let the surrounding temperature be $T^0$ C

Average T= $\frac{(50+45)}{2}$

$\frac{95}{2}$ = 47.5

$\frac{(50-45)}{5}=1^0$ C/m

Rate of fall of temperature = 5

From Newton's law

$1^0$ C/mm = bA $\times$ T

$\Rightarrow bA=\frac{1}{T}=\frac{1}{(47.5-t)}$

In 2nd case,
Average temperature

= $\frac{(40+45)}{2}=\frac{85}{2}$ = 42.5

Average temperaure difference from surrounding

T = 42.5 -t

Rate of fall of temperature

= $\frac{(45-40)}{8}={\left(\frac{5}{8}\right)}^0$ C/m

From Newton's law

$\frac{5}{8}$= bAT

$\frac{5}{8}=\frac{42.5-t}{47.5-t}$

Hence, t = ${34.1}^0$C