Discussion Forum

A body cools from $70^{0}$ C to $40^{0}$ C in 5 min. Calculate the time it takes to cool from $60^{0}$ C to $30^{0}$C . The temperature of the surrounding is $20^{0}$ C,

Answer:

Explanation:

 Body cools from 70° C to 40°C in 5 minute, hence , $T_{1}=70^{0}C, T_{2}=40^{0}C$ and t=5 minute 

Temperature of surrounding , $T_{0}=20^{0}C$

 By Newton's law of cooling,

$\frac{T_{1}-T_{2}}{t}=k\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right]$

or  $\frac{70-40}{5}=k\left[\frac{70+40}{2}-20\right]\Rightarrow k= \frac{6}{35}$  ...........(i)

 Again, let body cools from $60^{0}$C  to $30^{0} C$ in time t minutes.

 i.e, $T_{1}'=60^{0} C$ and $T_{2}'=30^{0}C$,

 $\therefore$   By Newton's law of cooling

$\frac{T_{1}'-T_{2}'}{t}=k\left[\frac{T_{1}'+T_{2}'}{2}-T_{0}\right]$

$\frac{60-30}{t}=\frac{6}{35}\left[\frac{60+30}{2}-20\right]$        [ $\because$ From eq.(i)]

 $\frac{30}{t}=\frac{6}{35}[25] \Rightarrow$ t=7 minutes

 Hence, option (b) is correct