Question

A hot body is placed in cold surroundings. Its rate of cooling is $3^o$C per minute when its temperature is ${70}^o$C and ${1.5}^o$C per minute when its temperature is ${50}^o$C. Its rate of cooling when temperature is ${40}^o$C is

• A. 0.25${}^o$C/min
• B. 0.5${}^o$C/min
• C. 0.75${}^o$C/min
• D. 1${}^o$C/min

Answer 1

The correct option is C 0.75${}^o$C/min

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

putting the given values of the two situations in the formula we get:-

$3=-k(70-{\theta }_0)$ - eq.1

$1.5=-k(50-{\theta }_0)$ - eq.2

Now, solving eq.1 and eq.2 we have

${\theta }_0=30{}^{0}C$ and$-k=\frac{3}{40}$

Now solving using the above values for ${40}^{0}C$, we have

$\frac{d\theta }{dt}=\frac{3}{40}(40-30)$

$\frac{d\theta }{dt}=\frac{3}{40}\left(10\right)$

$\frac{d\theta }{dt}=\frac{3}{4}={0.75}^{0}C/min$