Consider a classroom that is roughly `5mxx10mxx3m`. Initially `T=27^(@)`C and p=1 atm. There are 50 people in a insulated class losing energy to the room at the average rate of 150 Watt per person. How long can they remain in class if the body temperature is `42^(@)` C and person feels uncomfortable above this temperature. Heat capacity of air=`(7//2)R`

Options

A. `4.34` minutes
B. `5.91` minutes
C. `6.86` minutes
D. `7.79` minutes

Answer

To solve the problem step by step, we will follow the given data and apply the relevant thermodynamic principles.

### Step 1: Gather the Given Data
- Dimensions of the classroom: \(5 \, m \times 10 \, m \times 3 \, m\)
- Initial temperature of the room, \(T_i = 27^\circ C\)
- Final uncomfortable temperature, \(T_f = 42^\circ C\)
- Number of people, \(N = 50\)
- Rate of energy loss per person, \(P = 150 \, W\)
- Heat capacity of air, \(C_p = \frac{7}{2} R\)
- Pressure, \(P = 1 \, atm\)

### Step 2: Calculate the Volume of the Classroom
The volume \(V\) of the classroom can be calculated as:
\[ V = 5 \, m \times 10 \, m \times 3 \, m = 150 \, m^3 \]

### Step 3: Convert Volume to Liters
Since \(1 \, m^3 = 1000 \, L\):
\[ V = 150 \, m^3 \times 1000 \, L/m^3 = 150000 \, L \]

### Step 4: Calculate the Moles of Air in the Classroom
Using the Ideal Gas Law:
\[ PV = nRT \]
Where:
- \(P = 1 \, atm = 101.325 \, kPa\)
- \(R = 0.0821 \, L \cdot atm/(K \cdot mol)\)
- \(T = 27^\circ C + 273.15 = 300.15 \, K\)

Rearranging for \(n\):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1 \, atm)(150000 \, L)}{(0.0821 \, L \cdot atm/(K \cdot mol))(300.15 \, K)} \approx 6070.5 \, mol \]

### Step 5: Calculate the Heat Capacity of Air
Using the given heat capacity:
\[ C_p = \frac{7}{2} R = \frac{7}{2} \times 8.314 \, J/(mol \cdot K) \approx 29.3 \, J/(mol \cdot K) \]

### Step 6: Calculate the Total Heat Required to Raise the Temperature
The change in temperature \(\Delta T\) is:
\[ \Delta T = T_f - T_i = 42^\circ C - 27^\circ C = 15 \, K \]
The total heat \(Q\) required is:
\[ Q = n \cdot C_p \cdot \Delta T \]
Substituting the values:
\[ Q = 6070.5 \, mol \cdot 29.3 \, J/(mol \cdot K) \cdot 15 \, K \approx 267,000 \, J \]

### Step 7: Calculate the Total Power Loss
The total power loss from all individuals is:
\[ P_{total} = N \cdot P = 50 \cdot 150 \, W = 7500 \, W \]

### Step 8: Calculate the Time to Reach Uncomfortable Temperature
Using the formula:
\[ Q = P_{total} \cdot t \]
Rearranging for \(t\):
\[ t = \frac{Q}{P_{total}} = \frac{267000 \, J}{7500 \, W} \approx 35.6 \, s \]

### Step 9: Convert Time to Minutes
To convert seconds to minutes:
\[ t \approx \frac{35.6 \, s}{60} \approx 0.593 \, minutes \]

### Final Answer
The time they can remain in the classroom before feeling uncomfortable is approximately **0.593 minutes**.