Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K. A heater transfers energy at a constant rate of 12 W to the nitrogen. A piston maintains the pressure at 1.0 x 10^6 Pa during the heating process. The nitrogen is heated from 70 K and is completely turned into a gas after 890 s. Calculate the specific heat capacity of liquid nitrogen. Give an appropriate unit for your answer. specific latent heat of vaporisation of nitrogen = 2.0 x 10^5 J kg^-1 boiling point of nitrogen = 77 K [5 marks] The work done by the nitrogen in the cylinder when expanding due to a change of state is X. The energy required to change the state of the nitrogen from a liquid to a gas is Y. Deduce which is greater, X or Y. density of liquid nitrogen at its boiling temperature = 810 kg m^-3 density of nitrogen gas at its boiling temperature = 3.8 kg m^-3

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Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K - AQA - A-Level Physics - Question 1 - 2019 - Paper 2

Question 1

Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K. A heater transfers energy at a constant rate of 12 W t... show full transcript

Worked Solution & Example Answer:Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K - AQA - A-Level Physics - Question 1 - 2019 - Paper 2

Step 1

Calculate the specific heat capacity of liquid nitrogen.

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Answer

To find the specific heat capacity of liquid nitrogen, we must first calculate the total energy supplied by the heater. This is given by:

$Q = P imes t$ where:

$P = 12 \, \text{W}$ (power of the heater)
$t = 890 \, \text{s}$ (time duration)

Thus,

$Q = 12 \, \text{W} \times 890 \, \text{s} = 10680 \, \text{J}$

Next, we can express the energy in terms of the mass of nitrogen and its specific heat capacity:

$Q = m c \Delta T$ where:

$m = 0.050 \, \text{kg}$
$c$ is the specific heat capacity we want to find.
$\Delta T = 77 \, \text{K} - 70 \, \text{K} = 7 \, \text{K}$

Rearranging the equation for $c$ , we have:

$c = \frac{Q}{m \Delta T}$

Substituting the values:

$c = \frac{10680 \, \text{J}}{0.050 \, \text{kg} \times 7 \, \text{K}} = 30600 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$

Therefore, the specific heat capacity of liquid nitrogen is approximately $30600 \, \text{J kg}^{-1} \text{K}^{-1}$ .

Step 2

The work done by the nitrogen in the cylinder when expanding due to a change of state is X.

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Answer

To find the work done during the expansion of the nitrogen gas, we can use the formula:

$W = P \Delta V$

Where:

$P = 1.0 \times 10^6 \, \text{Pa}$ (constant pressure)

To find the volume change, we will compare the densities of nitrogen in its liquid and gaseous states:

Density of liquid nitrogen: $\rho_{liquid} = 810 \, \text{kg m}^{-3}$
Density of nitrogen gas: $\rho_{gas} = 3.8 \, \text{kg m}^{-3}$

Using the equation for density ( $\rho = \frac{m}{V}$ ), we can find the respective volumes:

For the liquid nitrogen:

$V_{liquid} = \frac{m}{\rho_{liquid}} = \frac{0.050 \, \text{kg}}{810 \, \text{kg m}^{-3}} \approx 6.17 \times 10^{-5} \, \text{m}^3$

For the nitrogen gas:

$V_{gas} = \frac{m}{\rho_{gas}} = \frac{0.050 \, \text{kg}}{3.8 \, \text{kg m}^{-3}} \approx 0.01316 \, \text{m}^3$

Thus, the change in volume ( $\Delta V$ ) is:

$\Delta V = V_{gas} - V_{liquid} \approx 0.01316 \, \text{m}^3 - 6.17 \times 10^{-5} \, \text{m}^3 \approx 0.01310 \, \text{m}^3$

Calculating the work done:

$W = 1.0 \times 10^6 \, \text{Pa} \times 0.01310 \, \text{m}^3 \approx 13100 \, \text{J}$

Thus, the work done is $W \approx 13100 \, \text{J}$ .

Step 3

The energy required to change the state of the nitrogen from a liquid to a gas is Y.

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Answer

The energy required to change the state of nitrogen from a liquid to a gas can be calculated using the specific latent heat of vaporization:

$Y = m L$ where:

$m = 0.050 \, \text{kg}$ (mass of the nitrogen)
$L = 2.0 \times 10^5 \, \text{J kg}^{-1}$ (latent heat of vaporization)

Thus,

$Y = 0.050 \, \text{kg} \times 2.0 \times 10^5 \, \text{J kg}^{-1} = 10000 \, \text{J}$

Therefore, the energy required to change the state of nitrogen is $Y \approx 10000 \, \text{J}$ .

Step 4

Deduce which is greater, X or Y.

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Answer

From our calculations:

The work done by nitrogen ( $X \approx 13100 \, \text{J}$ )
The energy required to change the state of nitrogen ( $Y \approx 10000 \, \text{J}$ )

We can see that:

$X > Y$

Thus, the work done by the nitrogen in the cylinder when expanding is greater than the energy required to change its state.

Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K - AQA - A-Level Physics - Question 1 - 2019 - Paper 2

Worked Solution & Example Answer:Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K - AQA - A-Level Physics - Question 1 - 2019 - Paper 2

Calculate the specific heat capacity of liquid nitrogen.

The work done by the nitrogen in the cylinder when expanding due to a change of state is X.

The energy required to change the state of the nitrogen from a liquid to a gas is Y.

Deduce which is greater, X or Y.

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