When a body is placed in surroundings at a constant temperature of `20^(@)C` and heated by a 10W heater, its temperature remains constant at `40^(@)C`. If the temperature of the body is now raised from `20^(@)C` to `80^(@)C` in 5 min at a uniform rate, the total heat it will lose to the surroundings will be

Options

A. 3000 J
B. 3600J
C. 4500J
D. 5400J

Answer

To solve the problem, we will apply Newton's Law of Cooling and calculate the total heat lost by the body when its temperature is raised from `20°C` to `80°C` in `5 minutes`.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body is initially at `20°C` and is heated to `40°C` with a `10W` heater, maintaining a constant temperature of `40°C` due to heat loss to the surroundings at `20°C`.
- We need to find the total heat lost when the body is uniformly heated from `20°C` to `80°C` in `5 minutes`.

2. **Determine the Rate Constant (K)**:
- According to Newton's Law of Cooling, the rate of heat loss \( \frac{dq}{dt} \) is proportional to the temperature difference between the body and the surroundings:
\[ \frac{dq}{dt} = K (\theta - \theta_0) \]
- Here, \( \theta = 40°C \) and \( \theta_0 = 20°C \).
- The power of the heater is `10W`, which means it provides `10J` of energy per second.
- At equilibrium, the heat gained by the body equals the heat lost:
\[ 10 = K (40 - 20) \]
\[ 10 = K \times 20 \]
\[ K = \frac{10}{20} = 0.5 \, \text{W/°C} \]

3. **Calculate the Total Heat Loss During Heating**:
- The body is heated from `20°C` to `80°C` in `5 minutes` (which is `300 seconds`).
- The temperature difference at any time \( t \) can be expressed as:
\[ \theta(t) = 20 + \frac{(80 - 20)}{300} \times t = 20 + \frac{60}{300} \times t = 20 + 0.2t \]
- The rate of heat loss at any time \( t \) is:
\[ \frac{dq}{dt} = K (\theta(t) - 20) = 0.5 \times (20 + 0.2t - 20) = 0.5 \times 0.2t = 0.1t \]

4. **Integrate to Find Total Heat Loss**:
- To find the total heat lost \( Q \) over the time interval from `0` to `300 seconds`, we integrate:
\[ Q = \int_0^{300} \frac{dq}{dt} \, dt = \int_0^{300} 0.1t \, dt \]
- Evaluating the integral:
\[ Q = 0.1 \left[ \frac{t^2}{2} \right]_0^{300} = 0.1 \times \frac{300^2}{2} = 0.1 \times 45000 = 4500 \, \text{J} \]

### Final Answer:
The total heat lost to the surroundings when the body is raised from `20°C` to `80°C` in `5 minutes` is **4500 Joules**.