Question

A body cools from 50${}^o$C to 45${}^o$C in 5 min and to 40${}^o$C in another 8 min. The temperature of the surrounding is

• A. 34${}^o$
• B. 30${}^o$
• C. 37${}^o$
• D. 43${}^o$

Answer 1

Answer: (C)

Newton's cooling law $\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$

putting the given values of the situation in the formula we get:-

$\frac{(50-45)}{5}=-k(50-T)$

$=>1=-k(50-T)$ -----(1)

Now, cooling body at takes 8 min to reach $40{}^{0}C$ form $45{}^{0}C$

$\frac{(45-40)}{8}=-k(45-T)$

$\frac{5}{8}=-k(45-T)$ ------(2)

Dividing the two equations, we get:

$8(45-T)=5(50-T)$

$360-8T=250-5T$

$T=110/3={36.67}^{0}C$