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A perfectly insulated flask contains a sample of metal M at a temperature of -10 °C - AQA - A-Level Physics - Question 1 - 2020 - Paper 2

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A perfectly insulated flask contains a sample of metal M at a temperature of -10 °C. Figure 1 shows how the temperature of the sample changes when energy is transfe... show full transcript

Worked Solution & Example Answer:A perfectly insulated flask contains a sample of metal M at a temperature of -10 °C - AQA - A-Level Physics - Question 1 - 2020 - Paper 2

Step 1

State the melting temperature of M.

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Answer

The melting temperature of M is 28 °C.

Step 2

Explain how the energy transferred to the sample changes the arrangement of the atoms during the time interval t_A to t_B.

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Answer

The energy transferred causes the atoms to vibrate more vigorously, which reduces the number of nearest atomic neighbors. This means the bonds that hold the atoms together start to break as they gain more energy, transitioning from a solid state towards a liquid state.

Step 3

State what happens to the potential energy of the atoms and to the kinetic energy of the atoms during the time interval t_A to t_B.

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Answer

During the interval t_A to t_B, the total (or mean) kinetic energy remains constant while the total (or mean) potential energy increases due to the breaking of bonds.

Step 4

Describe how the motion of the atoms changes during the time interval t_B to t_C.

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Answer

During the time interval t_B to t_C, the motion of the atoms changes from being more fixed in position to becoming more fluid as they transition into the liquid state. The atoms gain energy, allowing for increased movement.

Step 5

Determine the specific heat capacity of M when in the liquid state.

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Answer

The specific heat capacity of M in the liquid state can be calculated using the formula:

Q=mchetaQ = mc heta

Where:

  • QQ is the heat supplied,
  • mm is the mass,
  • cc is the specific heat capacity,
  • θ\theta is the change in temperature.

Given the values:

  • m=0.25extkgm = 0.25 ext{ kg},
  • Q=35extW×60 s=2100extJQ = 35 ext{ W} \times 60 \text{ s} = 2100 ext{ J},
  • Change in temperature from -10 °C to 30 °C = 40 °C,

Thus:

2100=0.25×c×402100 = 0.25 \times c \times 40

From which:

c=21000.25×40=210extJkg1 °C1c = \frac{2100}{0.25 \times 40} = 210 ext{ J kg}^{-1} \text{ °C}^{-1}

Therefore, the specific heat capacity of M is 210 J kg⁻¹ °C⁻¹.

Step 6

Identify M.

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Answer

Based on the provided latent heats of fusion in Table 1, M corresponds to Gallium, which has a latent heat of fusion of 80 kJ kg⁻¹.

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