Question

A body cools in $7$ min from ${60}^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$. What will be its temperature after the next $7$ min ?
The temperature of surroundings is ${10}^{\circ }\text{C}$. [Assume Newton's Law of cooling is applicable]

• A. $T={32}^{\circ }\text{C}$
• B. $T={28}^{\circ }\text{C}$
• C. $T={24}^{\circ }\text{C}$
• D. $T={36}^{\circ }\text{C}$

Answer 1

The correct option is B $T={28}^{\circ }\text{C}$

Using Newton's Law of cooling.
$\left(\frac{T_1-T_2}{t}\right)=K[\left(\frac{T_1+T_2}{2}\right)-T_0]$

$\Rightarrow \left(\frac{60-40}{7\times 60}\right)=K[\left(\frac{60+40}{2}\right)-10]$

$\Rightarrow \left(\frac{20}{7\times 60}\right)=K\times \mathrm{40................}\text{(I)}$

Let the final temperature by $T^{\circ }\text{C}$

$\left(\frac{40-T}{7\times 60}\right)=K[\left(\frac{40+T}{2}\right)-10]$

$\left(\frac{40-T}{7\times 60}\right)=K\left[\frac{20+T}{2}\right].....\text{(II)}$

dividing Eq.(I) to Eq.(II)

$\Rightarrow 160-4T=20+T$
$\Rightarrow T={28}^{\circ }\text{C}$
Hence, option (b) is the correct answer.