A body having mass 2 Kg 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.
When the body temperature has reached `36^(@)C`. it is heated again so that it reaches to `40^(@)C` in `10` minutes. Assume that the rate of loss of heat at `38^(@)C` is the average rate of loss for the given time . The total heat required from a heater by the body is :

Options

A. `7.2J`
B. `0.728 J`
C. `16J`
D. `32J`

Explanation

To solve the problem, we will follow the steps outlined below:

### Step 1: Understand the problem
We have a body with a mass of 2 kg that cools from an initial temperature of 40°C to a final temperature of 36°C in 10 minutes. The surrounding temperature is constant at 30°C. We need to calculate the total heat required to heat the body back from 36°C to 40°C in 10 minutes.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of heat loss is proportional to the temperature difference between the body and its surroundings.

The formula is:
\[ \frac{dQ}{dt} = -k(T - T_s) \]
where:
- \(dQ\) is the heat lost,
- \(dt\) is the time interval,
- \(k\) is the cooling constant,
- \(T\) is the temperature of the body,
- \(T_s\) is the surrounding temperature.

### Step 3: Calculate the cooling constant \(k\)
The body cools from 40°C to 36°C in 10 minutes. The average temperature during this time can be approximated as:
\[ T_{avg} = \frac{40 + 36}{2} = 38°C \]
The temperature difference with the surroundings is:
\[ \Delta T = T_{avg} - T_s = 38 - 30 = 8°C \]
Using the formula for Newton's Law of Cooling, we can express it as:
\[ \frac{(40 - 30) - (36 - 30)}{10} = k \cdot (38 - 30) \]
Substituting the values:
\[ \frac{10 - 6}{10} = k \cdot 8 \]
This simplifies to:
\[ \frac{4}{10} = 8k \implies k = \frac{1}{20} \]

### Step 4: Calculate the rate of heat loss
The rate of heat loss at 38°C can be calculated using:
\[ \frac{dQ}{dt} = k \cdot (T - T_s) = \frac{1}{20} \cdot (38 - 30) = \frac{1}{20} \cdot 8 = \frac{8}{20} = 0.4 \text{ J/min} \]

### Step 5: Calculate total heat lost in 10 minutes
Total heat lost in 10 minutes is:
\[ Q_{lost} = \frac{dQ}{dt} \cdot t = 0.4 \text{ J/min} \cdot 10 \text{ min} = 4 \text{ J} \]

### Step 6: Calculate the heat required to raise the temperature from 36°C to 40°C
The heat required to raise the temperature of the body back to 40°C can be calculated using the formula:
\[ Q_{required} = m \cdot C \cdot \Delta T \]
where:
- \(m = 2 \text{ kg}\),
- \(C = 2 \text{ J/°C}\),
- \(\Delta T = 40 - 36 = 4°C\).

Substituting the values:
\[ Q_{required} = 2 \cdot 2 \cdot 4 = 16 \text{ J} \]

### Final Answer
The total heat required from the heater by the body is **16 Joules**.

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