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A student heats 0.20 kg of water in a beaker until it evaporates into steam - OCR Gateway - GCSE Physics - Question 19 - 2021 - Paper 1

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Question 19

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A student heats 0.20 kg of water in a beaker until it evaporates into steam. The starting temperature of the water is 20 °C. (a) (i) The specific heat capacity of w... show full transcript

Worked Solution & Example Answer:A student heats 0.20 kg of water in a beaker until it evaporates into steam - OCR Gateway - GCSE Physics - Question 19 - 2021 - Paper 1

Step 1

Calculate the energy needed to raise the temperature of the water to 100 °C.

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Answer

To calculate the energy required to raise the temperature, we can use the formula:

Q=mcΔTQ = mc\Delta T

where:

  • QQ is the energy (in joules),
  • mm is the mass of the water (0.20 kg),
  • cc is the specific heat capacity (4200 J/kg°C), and
  • ΔT\Delta T is the change in temperature (final temperature - initial temperature).

First, we need to find the change in temperature:

ΔT=100°C20°C=80°C\Delta T = 100 °C - 20 °C = 80 °C

Now, substituting the values into the equation:

Q=0.20 kg×4200 J/kg°C×80°CQ = 0.20 \text{ kg} \times 4200 \text{ J/kg°C} \times 80 °C

Calculating this gives:

Q=0.20×4200×80=67200 JQ = 0.20 \times 4200 \times 80 = 67200 \text{ J}

Therefore, the energy needed is 67200 J.

Step 2

Calculate the energy needed to turn all the water at 100 °C into steam.

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Answer

To calculate the energy needed to turn all the water at 100 °C into steam, we will use the formula:

Q=mLQ = mL

where:

  • QQ is the energy (in joules),
  • mm is the mass of the water (0.20 kg), and
  • LL is the specific latent heat of vaporisation (2260000 J/kg).

Substituting the values into the equation:

Q=0.20 kg×2260000 J/kgQ = 0.20 \text{ kg} \times 2260000 \text{ J/kg}

Calculating this gives:

Q=0.20×2260000=452000 JQ = 0.20 \times 2260000 = 452000 \text{ J}

Therefore, the energy needed to turn all the water into steam is 452000 J.

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