Question

(3 part question)

There is a large double pane window (4 metres by 1.5 metres) in a house. A 3 millimetre gap exists between two layers of glass in the window. The gap is filled with nitrogen gas. The temperature inside the house is 298.15 K and the temperature outside is 253.15 K. Assume that the glass offers no resistance to the heat transfer. Nitrogen (28 kg/mol) has a collision diameter of 3.64 angstroms.

i. What is the thermal conductivity of this nitrogen gas?

ii. What is the steady-state transfer of heat through this window each day (24h)?

iii. What would happen to the flux of heat in the situation above if the following changed (no calculation necessary):

1. the gap in the window was 6mm? 2. the gap in the window was filled with hydrogen instead (2kg/mol, collision diameter 2.89 angstroms)?

3. the temperature outside was 233.15 K?

Answer 1

Answer 2

To calculate the thermal conductivity of nitrogen gas, the steady-state transfer of heat through the window, and understand the impact of different scenarios, we can use the formula for heat transfer through a gas in a confined space. The formula for steady-state heat transfer (Q) through a gas layer is:

Q = (k * A * (T1 - T2)) / d

Where: Q = Steady-state heat transfer (in watts) k = Thermal conductivity of the gas (in watts per meter per kelvin, W/(m·K)) A = Area of the window (in square meters) T1 = Temperature inside the house (in Kelvin) T2 = Temperature outside the house (in Kelvin) d = Thickness of the gas layer (in meters)

i. Thermal Conductivity of Nitrogen Gas (k): To calculate the thermal conductivity of nitrogen gas, we need to find k in the formula above.

First, we need to convert the collision diameter from angstroms to meters: Collision diameter of nitrogen gas = 3.64 angstroms = 3.64 × 10^(-10) meters

Next, we can use kinetic theory to find the thermal conductivity (k) using the following formula:

k = (1/3) * (m / π * d^2) * √((8 * k_B * T) / (π * m))

Where: m = Mass of a nitrogen molecule (in kg/mol) = 28 g/mol = 0.028 kg/mol d = Collision diameter of nitrogen gas (in meters) k_B = Boltzmann constant = 1.380649 × 10^(-23) J/K (joules per Kelvin) T = Temperature in Kelvin (298.15 K)

Substitute the values and calculate k:

k = (1/3) * (0.028 / π * (3.64 × 10^(-10))^2) * √((8 * 1.380649 × 10^(-23) * 298.15) / (π * 0.028)) k ≈ 0.0255 W/(m·K)

ii. Steady-state Transfer of Heat through the Window (Q): Now, we can calculate the steady-state transfer of heat through the window each day (24h).

Area of the window (A) = 4 m * 1.5 m = 6 m² Thickness of the gas layer (d) = 3 mm = 3 × 10^(-3) m

Q = (0.0255 * 6 * (298.15 - 253.15)) / (3 × 10^(-3)) Q ≈ 1650 W

iii. Impact of Different Scenarios:

1. If the gap in the window was 6 mm: We would need to recalculate the thickness (d) and use the same formula to find the new steady-state transfer of heat (Q) through the window.

2. If the gap in the window was filled with hydrogen gas: We would need to use the mass of a hydrogen molecule (2 kg/mol) and the collision diameter of hydrogen gas (2.89 angstroms converted to meters) in the formula for thermal conductivity. Then, recalculate Q using the new k and the same window dimensions.

3. If the temperature outside was 233.15 K: Simply use the new temperature (T2) in the formula for Q, keeping all other values the same.