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A student measures the density of glass - Edexcel - GCSE Physics - Question 7 - 2019 - Paper 1

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A student measures the density of glass. The student has - a bag of marbles, all made from the same type of glass - a weighing balance - a plastic measuring cylinde... show full transcript

Worked Solution & Example Answer:A student measures the density of glass - Edexcel - GCSE Physics - Question 7 - 2019 - Paper 1

Step 1

Describe how the student could find, as accurately as possible, the density of the glass used for the marbles.

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Answer

  1. Find the mass of the marbles:

    The student should use a weighing balance to accurately measure the total mass of the marbles in grams.

  2. Measure the water displacement:

    • The student should fill the plastic measuring cylinder with a known volume of water and take note of the initial water level.
    • Then, the marbles should be added to the cylinder which will cause the water level to rise. After adding the marbles, the student should again record the new water level to find the change in water level.
  3. Calculate the volume of water displaced:

    The volume of the water displaced, which is equal to the volume of the marbles, can be calculated by subtracting the initial water level from the new water level. This can be done in milliliters (mL) or cubic centimeters (cm³).

  4. Calculate the density:

    Using the formula for density:

    extDensity=massvolume ext{Density} = \frac{\text{mass}}{\text{volume}}

    The student will divide the mass of the marbles by the volume of water displaced to find the density in g/cm³ or kg/m³.

Step 2

Calculate the temperature of the water before it was heated.

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Answer

To find the temperature of the water before it was heated, we can use the equation for thermal energy:

Q=mcΔTQ = m \cdot c \cdot \Delta T

Where:

  • QQ is the total thermal energy supplied (84,000 J)
  • mm is the mass of the water (0.25 kg)
  • cc is the specific heat capacity of water (4200 J/kg°C)
  • ΔT\Delta T is the change in temperature

Rearranging this equation gives:

ΔT=Qmc\Delta T = \frac{Q}{m \cdot c}

Substituting the known values:

ΔT=840000.25×4200\Delta T = \frac{84000}{0.25 \times 4200}

ΔT=80°C\Delta T = 80 °C

The initial temperature before heating was at room temperature, commonly around 20°C. Therefore:

Temperature before heating=20°C+ΔT=20°C+80°C=100°C\text{Temperature before heating} = 20°C + \Delta T = 20°C + 80°C = 100°C

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