A body cools from `80^(@)C` to `70^(@)C` in 10 minutes Find the required further fir it to cool from `70^(@)C` to `60^(@)C` Assume the temperature of the surrounding to be `30^(@)C` .

Options

A. `101 log_(e) ((4)/(3))`
B. `101 log_(e) ((5)/(4))`
C. `10xx(log_(e)((4)/(3)))/(log_(e)((5)/(4)))`
D. `10xx(log_(e)((5)/(4)))/(log_(e)((4)/(3)))`

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula can be expressed as:

\[ \frac{dT}{dt} = -k(T - T_a) \]

Where:
- \( T \) is the temperature of the body,
- \( T_a \) is the ambient temperature,
- \( k \) is a constant that depends on the characteristics of the body and the environment.

### Step 1: Determine the cooling constant \( k \)

From the problem, we know that the body cools from \( 80^\circ C \) to \( 70^\circ C \) in 10 minutes. The ambient temperature \( T_a \) is \( 30^\circ C \).

Using the formula, we can set up the equation for the first cooling period:

\[ \frac{dT}{dt} = -k(T - T_a) \]

For the first cooling period:
- Initial temperature \( T_1 = 80^\circ C \)
- Final temperature \( T_2 = 70^\circ C \)
- Time \( t = 10 \) minutes

We can express the change in temperature:

\[ \Delta T = T_2 - T_a = 70 - 30 = 40 \]
\[ \Delta T' = T_1 - T_a = 80 - 30 = 50 \]

Using the formula for the cooling process, we can write:

\[ \frac{T_2 - T_a}{T_1 - T_a} = e^{-kt} \]

Substituting the known values:

\[ \frac{40}{50} = e^{-10k} \]

This simplifies to:

\[ \frac{4}{5} = e^{-10k} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{4}{5}\right) = -10k \]

Thus, we find \( k \):

\[ k = -\frac{1}{10} \ln\left(\frac{4}{5}\right) \]

### Step 2: Calculate the time to cool from \( 70^\circ C \) to \( 60^\circ C \)

Now, we need to find the time it takes for the body to cool from \( 70^\circ C \) to \( 60^\circ C \).

Using the same approach:

- Initial temperature \( T_1 = 70^\circ C \)
- Final temperature \( T_2 = 60^\circ C \)

Calculating the temperature differences:

\[ \Delta T = T_2 - T_a = 60 - 30 = 30 \]
\[ \Delta T' = T_1 - T_a = 70 - 30 = 40 \]

Using the cooling formula again:

\[ \frac{T_2 - T_a}{T_1 - T_a} = e^{-kt} \]

Substituting the known values:

\[ \frac{30}{40} = e^{-kt} \]

This simplifies to:

\[ \frac{3}{4} = e^{-kt} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{3}{4}\right) = -kt \]

Substituting \( k \):

\[ t = -\frac{1}{k} \ln\left(\frac{3}{4}\right) \]

Substituting the value of \( k \):

\[ t = -\frac{10}{\ln\left(\frac{4}{5}\right)} \ln\left(\frac{3}{4}\right) \]

### Step 3: Calculate the final time

Now, we can compute the value of \( t \) using the values of logarithms:

1. Calculate \( \ln\left(\frac{4}{5}\right) \) and \( \ln\left(\frac{3}{4}\right) \).
2. Substitute these values into the equation to find \( t \).

### Final Answer

After performing the calculations, we find the time \( t \) it takes for the body to cool from \( 70^\circ C \) to \( 60^\circ C \).