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 10^5 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, we first need to calculate the total energy supplied to the nitrogen using the formula:

$E = P imes t$

where:

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

Substituting the values, we get:

$E = 12 \, W imes 890 \, s = 10680 \, J$

Next, we must calculate the energy required to convert the nitrogen from a liquid to a gas:

Determine the mass of nitrogen: $m = 0.050 \, kg$
The energy required to vaporize the nitrogen is given by:

$Q_{vap} = m imes L_v$

where:
- $L_v = 2.0 imes 10^5 \, J \, kg^{-1}$ (specific latent heat of vaporization)
So,

$Q_{vap} = 0.050 \, kg imes 2.0 imes 10^5 \, J \, kg^{-1} = 10000 \, J$

Now we can find the specific heat capacity $ext{c}$ using:

$Q_{total} = Q_{vap} + Q_{heating}$

Where:

$Q_{heating} = m imes c imes ext{temperature change}$

The temperature change from 70 K to 77 K is:

$ext{temperature change} = 77 \, K - 70 \, K = 7 \, K$

Thus,

$Q_{heating} = 0.050 \, kg imes c imes 7 \, K$

Now we have,

$10680 \, J = 10000 \, J + 0.050 \, kg imes c imes 7 \, K$

Rearranging to find $ext{c}$ :

$680 \, J = 0.050 \, kg imes c imes 7 \, K$

$c = rac{680 \, J}{0.050 \, kg imes 7 \, K} = 1944.29 \, J \, kg^{-1} \, K^{-1}$

Thus, the specific heat capacity of liquid nitrogen is approximately $1944.29 \, J \, kg^{-1} \, K^{-1}$ .

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

The work done by the nitrogen ( $W$ ) as it expands can be calculated using the formula:

$W = P imes ext{change in volume}$

However, we are told to deduce whether $X$ or $Y$ is greater, where $Y$ represents the energy required for the phase change. It's crucial to note that if the volume change during expansion is negligible, the work done $W$ will also be relatively small compared to the energy needed to change the state.

Therefore, without specific volume values, we deduce that:

$X < Y.$

This implies that the energy required to change the state from liquid to gas ( $Y$ ) is greater than the work done by the nitrogen during expansion ( $X$ ).

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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

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.

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