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# A body cools from a temperature $3T$ to $2T$ in 10 minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be:(A) $\dfrac{7}{4}T$(B) $\dfrac{3}{2}T$(C) $\dfrac{4}{3}T$(D) $T$

Last updated date: 20th Jun 2024
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Hint: To find the temperature of the body which is cooling with time we use Newton’s law of cooling. We have to assume that the temperature of the surroundings is constant. The body cools down by radiating out the heat energy contained in it.

Formula Used:
The formulae used in the solution are given here.
$\dfrac{{dT}}{{dt}} = k\left( {{T_t} - {T_s}} \right)$ where $k$ is Newton’s cooling constant, ${T_t}$ is the temperature of the body at any time $t$, ${T_s}$ is the temperature of the surrounding.

Complete Step by Step Solution: Newton’s law of cooling describes the rate at which an exposed body changes temperature through radiation which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small.
According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.
$\dfrac{{dT}}{{dt}} = k\left( {{T_t} - {T_s}} \right)$ where $k$ is Newton’s cooling constant, ${T_t}$ is the temperature of the body at any time $t$, ${T_s}$ is the temperature of the surrounding.
On simplifying this law, we get,
$\ln \left( {\dfrac{{{T_2} - {T_1}}}{{{T_1} - T}}} \right) = kt$
It has been given that a body cools from a temperature $3T$ to $2T$ in 10 minutes, when the room temperature is $T$ and Newton's law of cooling is applicable.
For the first 10 minutes,
$\ln \left( {\dfrac{{3T - {T_s}}}{{2T - T}}} \right) = k\left( {10} \right)$
$\Rightarrow \ln \dfrac{{2T}}{T} = 10k$
Thus, we can write, $\ln 2 = 10k$.—(i)
For the next 10 minutes,
Let the temperature at the end of 10 minutes be $T'$.
$\ln \left( {\dfrac{{2T - {T_s}}}{{T' - T}}} \right) = 10k$
$\Rightarrow \ln \left( {\dfrac{T}{{T' - T}}} \right) = 10k$-(ii)
Dividing equation (i) by (ii),
$\ln 2 = \ln \left( {\dfrac{T}{{T' - T}}} \right)$
$\dfrac{{\ln 2}}{{\ln \left( {\dfrac{T}{{T' - T}}} \right)}} = \dfrac{{10k}}{{10k}}$
On simplification, we get,
$\dfrac{T}{{T' - T}} = 2$
$\Rightarrow 2T' = T + 2T = 3T$
The solution of this equation will give us the temperature at the end of 10 minutes, $T' = \dfrac{3}{2}T$.

Hence, the correct answer is Option B.

Note: The limitations of Newton’s Law of Cooling are stated below-
1. The difference in temperature between the body and surroundings must be small.
2. The loss of heat from the body should be by radiation only.
3. The major limitation of Newton’s law of cooling is that the temperature of surroundings must remain constant during the cooling of the body.