Question

A body initially at ${60}^{\circ }$C cools to ${50}^{\circ }$C in 10 min. What will be it's temperature at the end of the next 10 min. if the roc temperature is ${20}^{\circ }$C.Assume newton's law of cooling.

• A. ${42.50}^{\circ }$C
• B. ${45}^{\circ }$C
• C. ${40.46}^{\circ }$C
• D. ${44.23}^{\circ }$C

Answer 1

The correct option is A ${42.50}^{\circ }$C

According newton's law of cooling,

$T=T_s+(T_o-T_s)e^{-kt}$

where, T= temp. at any time t

$T_o$=initial temp.

$T_s$=surrounding temp.

t=time

k=cooling constant

since body cools initially from 60$℃$ to 50$℃$ in 10 min.

so from above equation,

50=20+(60-20)$e^{-k\left(10\right)}$

$\Rightarrow \frac{3}{4}=e^{-10k}$ ................eqution(1)

we need to find the temp. at the end of next 10 min. i.e. 20 min. from start

so again using newton's law of cooling formula,

$T=T_s+(T_o-T_s)e^{-kt}$

$\Rightarrow$T=20+(60-20)$e^{-k\left(20\right)}$

$\Rightarrow$T=20+(40)${\left\{e^{-10k}\right\}}^2$ ..................equation (2)

from equation (1) and (2), we get

$\Rightarrow$T=20+(40)${\left(\frac{3}{4}\right)}^2$

$\Rightarrow$T=42.5$℃$