Question

For a liquid, it is allowed to cool down in a room whose temperature is $20$, it takes 5 min. to cool down from 50 to 46 Find out temperature of liquid after next 5 min.

Answer 1

Newton's law of cooling states that the rate of heat loss of body is directly prop. to difference b/w its own temperature and ambient temperature.

$\frac{T_t-T_{\infty }}{T_i-T_{\infty }}=e^{-t/i}$

$\frac{T-T_{\infty }}{T_i-T_{\infty }}=e^{-t/i}$

Where, $T\to$ temperature at time

$T_{\infty }\to$ ehmibient temp.

$T_i\to$ initial temp.

So, $\frac{46-20}{50-20}=e^{-10/i}$

for next $5$ min, $t=10$ min

$\frac{T-20}{50-20}=e^{-10/i}$

$\Rightarrow \frac{T-20}{30}=\left(\frac{46-20}{30}\right)\left(\frac{46-20}{30}\right)$

$\Rightarrow T-20=\frac{26\times 26}{30}$

$\Rightarrow T={42.53}^{0}c$