Question

A liquid takes 10 minutes to cool from ${80}^{o}C$ to ${50}^{o}C$ Calculate the time it takes to cool from ${60}^{0}C$ to ${30}^{0}C.$ The temperature of the surroundings is ${20}^{0}C.$

• A. 5 min
• B. 45 min
• C. 8 min
• D. 9 min

Answer 1

Answer: (D)

9 min

Here$,$

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

Final temperature $(T-f)=50°C$

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

$t=5min$

$A/C$ 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 $\left(T_i\right)=60°C$

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

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.