A body cools in a surrounding of constant temperature `30^(@)C`. Its heat capacity is `2J//^(@)C`. Initial temperature of the body is `40^(@)C`. Assume Newton's law of cooling is valid. The body cools to `36^(@)C` in `10` minutes.
In further `10` minutes it will cool from `36^(@)C` to :

Options

A. `34.8 ^(@)C`
B. `32.1 ^(@)C`
C. `32.8^(@)C`
D. `33.6^(@)C`

Explanation

To solve the problem of how much the body cools in the next 10 minutes after cooling from 36°C to a lower temperature, we will use Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Understand the problem
The body cools from an initial temperature of 40°C to 36°C in 10 minutes, and we need to find the temperature after another 10 minutes. The surrounding temperature is constant at 30°C.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the surrounding temperature. Mathematically, it can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_s) \]
where \( T_s \) is the surrounding temperature, \( T \) is the temperature of the body, and \( k \) is a constant.

### Step 3: Calculate the cooling constant \( k \)
In the first 10 minutes, the body cools from 40°C to 36°C:
- Initial temperature \( T_1 = 40°C \)
- Final temperature \( T_2 = 36°C \)
- Surrounding temperature \( T_s = 30°C \)
- Time \( t = 10 \) minutes

Using the average temperature during this period:
\[ T_{avg1} = \frac{T_1 + T_2}{2} = \frac{40 + 36}{2} = 38°C \]
The temperature difference is:
\[ \Delta T_1 = T_1 - T_s = 40 - 30 = 10°C \]

Using the formula for cooling:
\[ \Delta T_1 = k \cdot t \cdot (T_{avg1} - T_s) \]
Substituting the values:
\[ 10 = k \cdot 10 \cdot (38 - 30) \]
\[ 10 = k \cdot 10 \cdot 8 \]
\[ 10 = 80k \implies k = \frac{10}{80} = \frac{1}{8} \text{ per minute} \]

### Step 4: Calculate the temperature after the next 10 minutes
Now, we need to find the temperature after another 10 minutes, starting from 36°C.

Using the average temperature for the next period:
\[ T_{avg2} = \frac{T_2 + T_n}{2} = \frac{36 + T_n}{2} \]
The temperature difference now is:
\[ \Delta T_2 = T_2 - T_s = 36 - 30 = 6°C \]

Using the cooling equation again:
\[ \Delta T_2 = k \cdot t \cdot (T_{avg2} - T_s) \]
Substituting the values:
\[ 6 = \frac{1}{8} \cdot 10 \cdot \left(\frac{36 + T_n}{2} - 30\right) \]
Simplifying:
\[ 6 = \frac{10}{8} \cdot \left(\frac{36 + T_n}{2} - 30\right) \]
\[ 6 = \frac{5}{4} \cdot \left(\frac{36 + T_n}{2} - 30\right) \]
Multiplying both sides by 4:
\[ 24 = 5 \cdot \left(\frac{36 + T_n}{2} - 30\right) \]
Dividing by 5:
\[ \frac{24}{5} = \frac{36 + T_n}{2} - 30 \]
Adding 30 to both sides:
\[ \frac{24}{5} + 30 = \frac{36 + T_n}{2} \]
Converting 30 to a fraction:
\[ \frac{24}{5} + \frac{150}{5} = \frac{36 + T_n}{2} \]
\[ \frac{174}{5} = \frac{36 + T_n}{2} \]
Multiplying both sides by 2:
\[ \frac{348}{5} = 36 + T_n \]
Subtracting 36 (which is \(\frac{180}{5}\)):
\[ \frac{348}{5} - \frac{180}{5} = T_n \]
\[ T_n = \frac{168}{5} = 33.6°C \]

### Final Answer
The body will cool from 36°C to **33.6°C** in the next 10 minutes.

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