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 on, after how much time interval again, the engine will turn off

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 system described, set up the equations for heat transfer, and then integrate to find the time interval after which the engine will turn off.

### Step 1: Understand the system
The room temperature is controlled between 20°C and 25°C. The heat engine turns on when the temperature drops below 20°C and turns off when it exceeds 25°C. The power of the engine is given by \( P = 3KT \), where \( K \) is a constant and \( T \) is the room temperature.

### Step 2: Set up the heat transfer equations
The room loses heat at a rate of \( K(T - T_0) \), where \( T_0 \) is the outdoor temperature (10°C). The incoming heat from the engine is \( 3KT \). Therefore, the net heat retained in the room can be expressed as:
\[ \frac{dQ}{dt} = \text{Incoming heat} - \text{Outgoing heat} \]
\[ \frac{dQ}{dt} = 3KT - K(T - T_0) = 3KT - KT + KT_0 = 2KT + KT_0 \]

### Step 3: Relate heat retention to temperature change
The heat retained in the room is also related to the change in temperature by:
\[ \frac{dQ}{dt} = C \frac{dT}{dt} \]
where \( C \) is the heat capacity of the room. Thus, we can equate the two expressions:
\[ C \frac{dT}{dt} = K(2T + T_0) \]

### Step 4: Rearrange and integrate
Rearranging gives:
\[ \frac{dT}{2T + T_0} = \frac{K}{C} dt \]
Integrating both sides, we will set the limits for temperature from 20°C to 25°C and for time from 0 to \( t \):
\[ \int_{20}^{25} \frac{dT}{2T + T_0} = \frac{K}{C} \int_{0}^{t} dt \]

### Step 5: Perform the integration
The left-hand side integrates to:
\[ \frac{1}{2} \ln(2T + T_0) \bigg|_{20}^{25} = \frac{1}{2} \left( \ln(50) - \ln(40) \right) = \frac{1}{2} \ln\left(\frac{50}{40}\right) = \frac{1}{2} \ln\left(\frac{5}{4}\right) \]

The right-hand side gives:
\[ \frac{K}{C} t \]

### Step 6: Substitute known values
We know \( T_0 = 10°C \) and \( C/K = 750 \). Thus:
\[ \frac{K}{C} = \frac{1}{750} \]
Substituting into the equation:
\[ \frac{1}{2} \ln\left(\frac{5}{4}\right) = \frac{1}{750} t \]

### Step 7: Solve for \( t \)
Now, we can solve for \( t \):
\[ t = 750 \cdot \frac{1}{2} \ln\left(\frac{5}{4}\right) \]
Using \( \ln\left(\frac{5}{4}\right) \approx 0.2231 \):
\[ t = 750 \cdot 0.11155 \approx 83.66 \text{ seconds} \]

### Step 8: Convert to minutes
To convert seconds to minutes:
\[ t \approx \frac{83.66}{60} \approx 1.39 \text{ minutes} \]

### Final Answer
The engine will turn off after approximately **1.39 minutes**.