Question

A body cools from ${50.0}^{\circ }C$ to ${48}^{\circ }C$ in 5$s$ . How long will it take to cool from ${40.0}^{\circ }C$ to ${39}^{\circ }C$ ? Assume the temperature of surroundings to be ${30.0}^{\circ }C$ and Newton's law of cooling to be valid.

• A. 2.5$min$
• B. 10$min$
• C. 20$min$
• D. 9$min$

Answer 1

The correct option is D 9$min$

Here$,$

Initial temperature $\left(Ti\right)=80°C$

Final temperature $\left(Tf\right)=50°C$

Temperature of the surrounding $\left(To\right)=20°C$

$t=5min$

$A/C$ to Newton's law of cooling

Rate of cooling $\left(dT/dt\right)=K\left[\right(T_i+T_f)/2-T_o]$

$(T_f-T_i)/t=K\left[\right(80+50)/2-20]$

$(80-50)/5=K[65-20]$

$6=K\times 45$

$K=6/45=2/15$

in second condition$,$

initial temperature$\left(Ti\right)=60°C$

Final temperature $\left(Tf\right)=30°C$

Time taken for cooling is$t$

$A/C$ Newton's law of cooling

$(60-30)/t=2/15\left[\right(60+30)/2-20]$

$30/t=2/15\times 25$

$30/t=50/15=10/3$

$t=9min$

Hence,

option $\left(D\right)$ is correct answer.