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 × 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 × 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

A-Level AQA Physics Thermal Energy Transfer: Figure 1 shows a perfectly insul

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 first need to calculate the total energy supplied by the heater:

$P = 12 \, \text{W}, \quad t = 890 \, \text{s}$

The total energy ( Q) supplied is:

$Q = P \times t = 12 \times 890 = 10 680 \, \text{J}$

Since the nitrogen is heated from 70 K to its boiling point of 77 K, the temperature change (ΔT) is:

$\Delta T = 77 - 70 = 7 \ \, \text{K}$

Using the formula for specific heat capacity (c):

$Q = m \cdot c \cdot \Delta T$

Where:

m = mass of liquid nitrogen = 0.050 kg
Q = total energy = 10 680 J

Rearranging gives:

$c = \frac{Q}{m \cdot \Delta T} = \frac{10 680}{0.050 \cdot 7} = \frac{10 680}{0.35} = 30 600 \, \text{J kg}^{-1} \text{K}^{-1}$

Thus, the specific heat capacity of liquid nitrogen is approximately 30 600 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 energy required to change the state of the nitrogen from a liquid to a gas is given by the latent heat of vaporisation:

$Y = m \cdot L \, \text{where} \, L = 2.0 \times 10^5 J kg^{-1}$ Substituting the mass:

$Y = 0.050 \cdot 2.0 \times 10^5 = 10 000 \, \text{J}$

The work done during expansion (X) can be calculated using the formula:

$X = P \Delta V$

Here, the change in volume (ΔV) is due to the density changes:

Density of liquid nitrogen = 810 kg m^-3
Density of nitrogen gas = 3.8 kg m^-3

Calculating the volumes:

Volume of liquid nitrogen: $V_{liquid} = \frac{m}{density_{liquid}} = \frac{0.050}{810} = 6.17 \times 10^{-5} \, \text{m}^3$

Volume of nitrogen gas: $V_{gas} = \frac{m}{density_{gas}} = \frac{0.050}{3.8} = 0.01316 \, \text{m}^3$

Thus,

$\Delta V = V_{gas} - V_{liquid} = 0.01316 - 6.17 \times 10^{-5} \approx 0.0131 \, \text{m}^3$

Now using the pressure given: $X = (1.0 \times 10^5) \cdot (0.0131) \approx 1 310 \, \text{J}$

Comparing X and Y, we see: Y (10 000 J) is greater than X (1 310 J). Therefore, Y is greater than X.

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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