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A student investigated how the temperature of a lump of ice varied as the ice was heated - AQA - GCSE Physics - Question 11 - 2021 - Paper 1

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A student investigated how the temperature of a lump of ice varied as the ice was heated. The student recorded the temperature until the ice melted and then the wat... show full transcript

Worked Solution & Example Answer:A student investigated how the temperature of a lump of ice varied as the ice was heated - AQA - GCSE Physics - Question 11 - 2021 - Paper 1

Step 1

The specific heat capacity of ice is less than the specific heat capacity of water.

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Answer

In Figure 17, the initial temperature increase for ice is less steep than the subsequent increase for water after melting. This demonstrates that less energy is required to raise the temperature of ice compared to water, indicating its lower specific heat capacity.

Step 2

The specific latent heat of fusion of ice is less than the specific latent heat of vaporisation of water.

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Answer

Figure 17 shows a flat region during the melting of ice, which represents the heat added without a temperature change. However, the temperature of water increases significantly once it starts boiling, demonstrating that the latent heat of vaporisation is greater than that of fusion.

Step 3

Describe two ways the results of the experiment in Figure 17 would have been different.

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Answer

  1. The gradient on the temperature vs. time graph would be less steep if more thermal energy was transferred to the surroundings, indicating a longer time taken for the temperature to rise.

  2. The flat regions representing phase changes would be extended, showing a longer time required to change states due to the increased transfer of energy to the surroundings.

Step 4

Calculate the specific latent heat of vaporisation of water.

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Answer

To calculate the specific latent heat of vaporisation (L), use the formula: L=EmL = \frac{E}{m} where E is the energy transferred (69,000 J) and m is the mass of water (0.030 kg). Thus, L=69,0000.030=2,300,000 J/kgL = \frac{69,000}{0.030} = 2,300,000 \text{ J/kg}. The unit is J/kg.

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