Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as:
`6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)`
The energy of one mole of a photon of wave lenght is known as one Einstein.
A glucose fuel cell was designed to generate 4000 Watts which was used to heat of `5mxx 5mxx 3m` dimensions . The room is perfectly insulated and no heat flows out of the room Calculate the time needed to increase the temperature of the room by `20^(@)C` , if the room was initially at `10^(@)C` and 1 atm pressure . Assume that the air in the room behaves like an ideal gas and its molar heat capacity is 2.5R.

Answer

To solve the problem, we need to calculate the time required to increase the temperature of a room with given dimensions and conditions using the power generated by a glucose fuel cell. Here’s the step-by-step solution:

### Step 1: Calculate the volume of the room
The dimensions of the room are given as 5 m x 5 m x 3 m.

\[ \text{Volume} (V) = \text{length} \times \text{width} \times \text{height} = 5 \, \text{m} \times 5 \, \text{m} \times 3 \, \text{m} = 75 \, \text{m}^3 \]

### Step 2: Calculate the mass of air in the room
Assuming the density of air at 1 atm and 10°C is approximately 1.25 kg/m³, we can calculate the mass of air in the room.

\[ \text{Mass} (m) = \text{Density} \times \text{Volume} = 1.25 \, \text{kg/m}^3 \times 75 \, \text{m}^3 = 93.75 \, \text{kg} \]

### Step 3: Calculate the heat capacity of the air in the room
The molar heat capacity of air is given as \(2.5R\), where \(R\) is the universal gas constant (\(R = 8.314 \, \text{J/(mol K)}\)).

First, we need to find the number of moles of air in the room. The molar mass of air is approximately 29 g/mol, which is 0.029 kg/mol.

\[ \text{Number of moles} (n) = \frac{\text{Mass}}{\text{Molar mass}} = \frac{93.75 \, \text{kg}}{0.029 \, \text{kg/mol}} \approx 3224.14 \, \text{mol} \]

Now, we can calculate the total heat capacity (C) of the air in the room:

\[ C = n \times 2.5R = 3224.14 \, \text{mol} \times 2.5 \times 8.314 \, \text{J/(mol K)} \approx 67093.6445 \, \text{J/K} \]

### Step 4: Calculate the heat required to increase the temperature
The temperature change (\(\Delta T\)) is given as \(20°C\) (from \(10°C\) to \(30°C\)).

\[ Q = C \times \Delta T = 67093.6445 \, \text{J/K} \times 20 \, \text{K} \approx 1341872.89 \, \text{J} \]

### Step 5: Calculate the time required to heat the room
The power generated by the glucose fuel cell is given as 4000 Watts (W), which is equivalent to Joules per second (J/s).

Using the formula:

\[ \text{Time} (t) = \frac{Q}{\text{Power}} = \frac{1341872.89 \, \text{J}}{4000 \, \text{W}} \approx 335.47 \, \text{s} \]

### Final Answer
The time needed to increase the temperature of the room by \(20°C\) is approximately **335.47 seconds**.

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