Question

If a body cools from ${80}^o$C to ${50}^o$C at room temperature of ${25}^o$C in $30$ minutes, then the temperature of the body after $1$ hour is?

• A. ${36}^o$C
• B. ${38}^o$C
• C. ${36.36}^o$C
• D. ${39}^o$C

Answer 1

The correct option is A ${36}^o$C

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 $\left(\frac{dT}{dt}\right)=K[\frac{(Ti+Tf)}{2}-To]$

$\frac{(Tf-Ti)}{t}=K\left[\frac{(80+50)}{2-20}\right]$

$\frac{(80-50)}{5}=K[65-20]$

$6=K\times 45$

$K=\frac{6}{45}=\frac{2}{15}$

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

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

$\frac{30}{t}=\frac{2}{15}\times 25$

$\frac{30}{t}=\frac{50}{15}$

$=\frac{10}{3}$

$t=9min$