Question

A body originally at ${60}^{\circ }$ C cools down to ${40}^{\circ }$ C in 15 min when kept in air at a temperature of ${25}^{\circ }$ C. What will be the temperature of the body at the end of 30 min?

• A. 35.2${}^{\circ }$ C
• B. 31.5${}^{\circ }$ C
• C. 28.7${}^{\circ }$ C
• D. 15${}^{\circ }$ C

Answer 1

The correct option is B 31.5${}^{\circ }$ C

Let T : temperature at time 't'

$T_0:$ temperature of the surrounding (air)

Given, $T_0={25}^{\circ }C$

By Newton's law of cooling

$\frac{dT}{dt}=-K(T-T_0)$

{ K $\to$ constant of integration}

$\frac{dT}{(T-T_0)}=-Kdtintegrating,weget$

$ln(T-T_0)=-Kt+C$

T - $T_0=e^{-kt}{e}^c;$

{ C $\to$ constant of integration}

Given at t = 0, T = ${60}^{\circ }$

$\Rightarrow 60-25=e^{-o}{e}^c$

$e^c=35$

at t = 15 min

T = ${40}^{\circ }C$

40 - 25 = $e^{-15k}\times 35$

$e^{15K}=\frac{35}{15}=\frac{7}{3}$

at t = 30 min

T - 25 = $e^{-k30}35$

= $(e^{15k}{)}^{(-2)}\times 35={\left(\frac{7}{3}\right)}^{-2}\times 35$

T = 25 + $\frac{9}{49}\times 35$

= 31.429${}^{\circ }C$