A body cools from 50•C to 48•Celsius in 5sec.how long will it take to cool from 40•C to 39•C? Assume the temp. Of surrounding to be 30.0•c and Newton's law of cooling to be valid.?

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Introduction:
In this problem, we are given the initial temperature of a body (50°C), the final temperature (48°C), and the time it takes to cool from the initial to final temperature (5 seconds). We are also given the temperature of the surrounding environment (30.0°C) and the assumption that Newton's law of cooling is valid. We need to determine how long it will take for the body to cool from 40°C to 39°C.

Understanding Newton's Law of Cooling:
Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings. Mathematically, it can be represented as:

$\frac{dT}{dt} = -k(T - T_s)$

Where:
- $\frac{dT}{dt}$ is the rate of change of temperature with respect to time
- $k$ is the cooling constant
- $T$ is the temperature of the body
- $T_s$ is the temperature of the surroundings

Finding the Cooling Constant:
To find the cooling constant, we can use the given information that the body cools from 50°C to 48°C in 5 seconds. Substituting these values into Newton's law of cooling, we can solve for $k$:

$-k(50 - 30) = \frac{48 - 30}{5}$
$-20k = \frac{18}{5}$
$k = -\frac{18}{5} \div -20$
$k = \frac{9}{200}$

Cooling Time Calculation:
Now that we have the cooling constant, we can use Newton's law of cooling to determine the time it takes for the body to cool from 40°C to 39°C. Let's assume $t$ represents the time in seconds.

$\frac{dT}{dt} = -\frac{9}{200}(T - 30)$
$\frac{1}{T - 30}dT = -\frac{9}{200}dt$

Integrating both sides:

$\int_{39}^{40} \frac{1}{T - 30}dT = -\int_{0}^{t} \frac{9}{200}dt$
$\ln|T - 30|\Big|_{39}^{40} = -\frac{9}{200}t\Big|_{0}^{t}$
$\ln\left|\frac{40 - 30}{39 - 30}\right| = -\frac{9}{200}t$

Simplifying:

$\ln\left(\frac{4}{3}\right) = -\frac{9}{200}t$
$t = -\frac{200}{9} \cdot \ln\left(\frac{4}{3}\right)$

Conclusion:
Therefore, the body will take approximately $t = -\frac{200}{9} \cdot \ln\left(\frac{4}{3}\right)$ seconds to cool from 40°C to 39°C.

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