Question

A body cools in 7 minutes from 60 'C to 40 'C .What will be its temperature after the next 7 minutes. The temperature of surroundings is 10 'C

Answer 1

any comments welcome

Answer 2

Answer:

$28^\circ\rm C$

Explanation:

According to Newton's law of cooling, the change in temperature of a body $\rm \Delta T$ in time interval $\rm \Delta t$ is given by

$\rm \dfrac{\Delta T}{\Delta t}=K(T-T_s).$

where,

• T = $\rm \dfrac{Final\ temp. + Initial\ temp.}{2}.$

• $\rm T_s$ = temperature of the surrounding.

• K = a constant.

Given that the body cools in 7 minutes from $60^\circ \rm C$ to $40^\circ \rm C$,

$\rm \Delta T = 60-40\\\Delta t = 7\ minutes.$

The temperature of the surroundings, $\rm T_s = 10^\circ C$

Therefore,

$\rm \dfrac{60-40}{7}=K \left ( \dfrac{60+40}{2}-10 \right )\\\dfrac{20}{7}=K(40)\\\\\Rightarrow K = \dfrac{1}{14}.$

After next 7 minutes, let the temperature of the body be $\rm T'$, therefore, on using Newton's law of cooling,

$\rm \dfrac{40-T'}{7}=K\left(\dfrac{40+T'}{2}-10 \right )\\\dfrac{40-T'}{7}=\dfrac 1{14}\left(\dfrac{40+T'-20}{2} \right )\\40-T'=\dfrac{1}{4}(20+T')\\160-4T'=20+T'\\-4T'-T'=20-160\\-5T'=-140\\T'=\dfrac{140}{5}=28^\circ C.$