A body takes 10 minutes to cool from $60^\circ C$ to $50^\circ C$. The temperature of surroundings is constant at $25^\circ C$. Then, the temperature of the body after next 10 minutes will be approximately :

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A body takes 10 minutes to cool from $60^\circ C$ to $50^\circ C$. The temperature of surroundings is constant at $25^\circ C$. Then, the temperature of the body after next 10 minutes will be approximately :

$47^\circ C$

$43^\circ C$

$41^\circ C$

$45^\circ C$

The correct answer is

$43^\circ C$

Understanding Newton's Law of Cooling

This problem involves Newton's Law of Cooling, which describes how the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.

The integrated form of Newton's Law of Cooling is given by:

$ \ln\left(\frac{T_1 - T_s}{T_2 - T_s}\right) = kt $

Where:

$T_1$ is the initial temperature of the body.
$T_2$ is the final temperature of the body.
$T_s$ is the constant temperature of the surroundings.
$k$ is the cooling constant.
$t$ is the time taken for the temperature change.

Calculating the Cooling Constant (k)

We are given the first cooling interval:

$T_1 = 60^\circ C$
$T_2 = 50^\circ C$
$T_s = 25^\circ C$
$t = 10$ minutes

Plugging these values into the formula:

$ \ln\left(\frac{60 - 25}{50 - 25}\right) = k \times 10 $ $ \ln\left(\frac{35}{25}\right) = 10k $ $ \ln\left(\frac{7}{5}\right) = 10k $

So, the cooling constant $k$ is:

$ k = \frac{1}{10} \ln\left(\frac{7}{5}\right) $

Determining Temperature After Next 10 Minutes

Now, we need to find the temperature ($T_3$) after the next 10 minutes. The second interval starts from $T_2 = 50^\circ C$ and lasts for $t = 10$ minutes.

Initial temperature for this interval: $T_2 = 50^\circ C$
Final temperature for this interval: $T_3$ (unknown)
Surrounding temperature: $T_s = 25^\circ C$
Time interval: $t = 10$ minutes

Using the formula again:

$ \ln\left(\frac{T_2 - T_s}{T_3 - T_s}\right) = kt $

Substitute the value of $k$ and the known temperatures:

$ \ln\left(\frac{50 - 25}{T_3 - 25}\right) = \left(\frac{1}{10} \ln\left(\frac{7}{5}\right)\right) \times 10 $ $ \ln\left(\frac{25}{T_3 - 25}\right) = \ln\left(\frac{7}{5}\right) $

Equating the terms inside the logarithm:

$ \frac{25}{T_3 - 25} = \frac{7}{5} $

Solve for $T_3$:

$ 25 \times 5 = 7 \times (T_3 - 25) $ $ 125 = 7T_3 - 175 $ $ 7T_3 = 125 + 175 $ $ 7T_3 = 300 $ $ T_3 = \frac{300}{7} $ $ T_3 \approx 42.857^\circ C $

Final Temperature Approximation

The calculated temperature $T_3$ is approximately $42.857^\circ C$. Comparing this with the given options, the closest value is $43^\circ C$. Therefore, the temperature of the body after the next 10 minutes will be approximately $43^\circ C$.

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