Question 761: Thermodynamics & KTG

Question

An ordinary body cools from ' $4 	heta^{\prime}$ ' to ' $3 	heta^{\prime}$ ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is $	heta$)

Accepted Answer

According to Newton's law of cooling, the rate at which an object cools is proportional to the difference between its temperature and the ambient temperature. The formula is:
$ \frac{	heta_1 - 	heta_2}{t} = K \left[\frac{	heta_1 + 	heta_2}{2} - 	heta_0ight] $
where:
$	heta_0$ is the ambient (surrounding) temperature,
$	heta_1$ and $	heta_2$ are the initial and final temperatures of the body,
$K$ is the cooling constant,
$t$ is the time.
Applying this to the problem at hand, we have:
$ \begin{aligned} 	herefore \quad & \frac{4	heta - 3	heta}{t} = K \left[\frac{4	heta + 3	heta}{2} - 	hetaight] \ & \frac{	heta}{t} = K 	imes \frac{5	heta}{2} \quad	ext{.... (i)}\ & 	ext{Solving for $K$, we find: } K = \frac{2}{5t} \end{aligned} $
After another $t$ minutes, let the new temperature be $x$. Then:
$ \begin{array}{ll} 	herefore & \frac{3	heta - x}{t} = \frac{2}{5t} \left[\frac{3	heta + x}{2} - 	hetaight] \quad \ldots[	ext{using (i)}] \ 	herefore & 3	heta - x = \frac{3	heta + x - 2	heta}{5} \ 	herefore & 15	heta - 5x = x + 	heta \ 	herefore & 6x = 14	heta \ 	herefore & x = \frac{7	heta}{3} \end{array} $
Therefore, the temperature of the body after the next $t$ minutes is $\frac{7	heta}{3}$.

Answer

$\frac{9 	heta}{4}$

Answer

$\frac{2 	heta}{5}$

Answer

$\frac{5 	heta}{3}$

Answer

$\frac{7 	heta}{3}$

Thermodynamics & KTG

An ordinary body cools from ' 4 \theta^{\prime} ' to ' 3 \theta^{\prime} ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is \theta)

Choose the correct answer: