Imagine a system, that can keep the room temperature within a narrow range between `20^@C` to `25^@C`. The system includes a heat engine operating with variable power P = 3KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in temperature drops lower than `20^@C`, the engine turns on, when the temperature increase over `25^@`C, the engine turns off, room looses energy at a rate of `K(T - T_0), T_0` is the outdoor temperature. The heat capacity of the room is C.
Given `(T_0=10^@C, ln(3/2) =0.4 , ln(6/5)=0.18 , C/K`=750 SI-unit)
Suppose at t = 0, the engine turns off, after how much time interval, again, the engine will turn on

Options

A. 10 min
B. 5 min
C. 1.125 min
D. 2.25 min

Answer

To solve the problem step by step, we will analyze the given information and derive the time interval after which the heat engine will turn on again.

### Step 1: Understand the System
The system maintains room temperature between 20°C and 25°C. The heat engine operates with power \( P = 3KT \), where \( K \) is a constant, and \( T \) is the room temperature. The engine turns on below 20°C and turns off above 25°C.

### Step 2: Identify Initial Conditions
At \( t = 0 \), the engine is off, which means the room temperature is at 25°C. We need to find the time it takes for the temperature to drop to 20°C.

### Step 3: Write the Heat Loss Equation
The room loses energy at a rate given by:
\[ \frac{dQ}{dt} = -K(T - T_0) \]
where \( T_0 = 10°C \) (the outdoor temperature). Since the engine is off, the incoming heat is zero.

### Step 4: Relate Heat Loss to Temperature Change
Using the heat capacity \( C \), we can relate the heat loss to the change in temperature:
\[ \frac{dQ}{dt} = C \frac{dT}{dt} \]
Setting the two equations equal gives:
\[ C \frac{dT}{dt} = -K(T - T_0) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{dT}{dt} = -\frac{K}{C}(T - T_0) \]

### Step 6: Integrate the Equation
We will integrate this equation from \( T = 25°C \) to \( T = 20°C \):
\[ \int_{25}^{20} \frac{1}{T - T_0} dT = -\frac{K}{C} \int_{0}^{t} dt \]
This results in:
\[ -\frac{C}{K} \left[ \ln(T - T_0) \right]_{25}^{20} = t \]

### Step 7: Substitute Limits
Substituting the limits into the logarithmic expression:
\[ -\frac{C}{K} \left( \ln(20 - 10) - \ln(25 - 10) \right) = t \]
This simplifies to:
\[ -\frac{C}{K} \left( \ln(10) - \ln(15) \right) = t \]
Using properties of logarithms, this can be rewritten as:
\[ -\frac{C}{K} \ln\left(\frac{10}{15}\right) = t \]
Thus:
\[ t = \frac{C}{K} \ln\left(\frac{15}{10}\right) \]

### Step 8: Substitute Known Values
Given \( \frac{C}{K} = 750 \) and using \( \ln\left(\frac{15}{10}\right) = \ln(1.5) \):
\[ t = 750 \cdot \ln(1.5) \]
Using the approximation \( \ln(1.5) \approx 0.4 \):
\[ t = 750 \cdot 0.4 = 300 \text{ seconds} \]

### Step 9: Convert to Minutes
To convert seconds to minutes:
\[ \text{Time in minutes} = \frac{300}{60} = 5 \text{ minutes} \]

### Final Answer
The heat engine will turn on again after **5 minutes**.

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