Question

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

Answer 1

Hint : First we know the temperature is a measure of the average kinetic energy of the particles in an object. When temperature increases, the motion of these particles also increases. In other words, temperature determines the internal energy within a given system.
Using Newton’s law of cooling formula we find out the temperature of the liquid at the given time.

Complete Step By Step Answer:
Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings.
$\dfrac{{{T_t} - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \dfrac{t}{i}}}$
$\Rightarrow$ $\dfrac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \dfrac{t}{i}}}$ ----(1)
Where, $T$ is a temperature at time, ${T_\infty }$ is ambient temperature (Atmospheric temperature) and ${T_i}$ is the initial temperature. Further, $\dfrac{1}{i}$ refers to a cooling constant.
Given a room temperature ${T_\infty } = {20^o}$ $C$ and liquid takes $t = 5$ min to cool down from ${T_i} = {50^o}$ $C$ to $T = {46^o}$ $C$ .
Then the equation (1) becomes
$\dfrac{{46 - 20}}{{50 - 20}} = {e^{ - \dfrac{5}{i}}}$ ----(2)
The equation (1) becomes for next $5$ min where $t = 10$ min is
$\dfrac{{T - 20}}{{50 - 20}} = {e^{ - \dfrac{{10}}{i}}}$ ---(3)
The equation (3) can be rewritten as
$\dfrac{{t - 20}}{{50 - 20}} = {\left( {{e^{ - \dfrac{5}{i}}}} \right)^2}$ ---(4)
Using the equation (2) in the equation (4)
$\Rightarrow \dfrac{{T - 20}}{{30}} = {\left( {\dfrac{{46 - 20}}{{30}}} \right)^2}$
$\Rightarrow T - 20 = \dfrac{{26 \times 26}}{{30}}$
$\Rightarrow T = {42.53^0}$ $C$ .

Note :
Note that absolute zero is the temperature at which there is no molecular motion. The three main temperature scales are Celsius, Fahrenheit, and Kelvin. Temperature is measured with a thermometer or a calorimeter.