Question

A body cools from ${70}^{o}C$ to ${50}^{o}C$ in 5 minutes. The temperature of the surroundings is ${20}^{o}C$. Its temperature after the next 10 minutes is:

• A. ${25}^{o}C$
• B. ${35}^{o}C$
• C. ${38}^{o}C$
• D. ${45}^{o}C$

Answer 1

The correct option is C ${38}^{o}C$

$hA\left(T-T\infty \right)=mlp\frac{dT}{dT}$

$\frac{hA}{mlp}(T–T_{\infty })=\frac{dT}{dt}$

$T_{\infty }={20}^{\circ }$ $C(T–T_{\infty })=\frac{dT}{dt}$

$C(T–T_{\infty })dt=dT$

$\int Cdt=\int \frac{dT}{T-T_{\infty }}$

$C\times t=log(T–T_{\infty })+C_1$

At $t=0,T={70}^{\circ }$

$0=log(70–20)+C_1$

$C_1=-log50$

And at $t=5min,T={50}^{\circ }$

$5\times C=log(50–20)-log50$

$=log30–log50$

$C=\frac{1}{5}log\left(\frac{3}{5}\right)$

$\frac{t}{5}log\left(\frac{3}{5}\right)=log(T–T_{\infty })-log50$

$\frac{t}{5}log\left(\frac{3}{5}\right)=log\left(\frac{T–T_{\infty }}{50}\right)$

At $t=10min$

$\frac{10}{5}log\left(\frac{3}{5}\right)+log50=log(T-20)$

$log\frac{9}{25}+log50=log(T-20)$

$log(\frac{9}{15}\times 50)=log(T-20)$

$18=T-20$

$T={38}^{\circ }$