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2 (a) Describe, in terms of particles, two differences between a solid and a liquid of the same substance - Edexcel - GCSE Physics Combined Science - Question 2 - 2021 - Paper 1

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2 (a) Describe, in terms of particles, two differences between a solid and a liquid of the same substance. (b) Figure 3 shows the dimensions of a solid block of co... show full transcript

Worked Solution & Example Answer:2 (a) Describe, in terms of particles, two differences between a solid and a liquid of the same substance - Edexcel - GCSE Physics Combined Science - Question 2 - 2021 - Paper 1

Step 1

Describe, in terms of particles, two differences between a solid and a liquid of the same substance.

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Answer

  1. In a solid, particles are packed closely together, maintaining a fixed structure, while in a liquid, particles are more loosely arranged and can move freely.

  2. The particles in a solid vibrate in fixed positions, whereas the particles in a liquid can move past one another, allowing liquids to flow.

Step 2

Calculate the mass of the concrete block.

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Answer

To find the mass of the concrete block, we first need to calculate its volume.

Using the dimensions given in Figure 3:

  • Length = 1.5 m
  • Width = 1.0 m
  • Height = 0.20 m

The volume (V) can be calculated as:

V=extLengthimesextWidthimesextHeightV = ext{Length} imes ext{Width} imes ext{Height}

Substituting the values:

V=1.5imes1.0imes0.20=0.3extm3V = 1.5 imes 1.0 imes 0.20 = 0.3 ext{ m}^3

Now, we can use the density (ρ) of concrete to find the mass (m):

m=ρimesVm = ρ imes V

Substituting the values:

m=2100extkg/m3imes0.3extm3=630extkgm = 2100 ext{ kg/m}^3 imes 0.3 ext{ m}^3 = 630 ext{ kg}

Thus, the mass of the concrete block is 630 kg.

Step 3

State two practical ways to reduce heat loss from this shed.

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Answer

  1. Insulating the walls and roof to minimize heat transfer.

  2. Installing double-glazed windows to reduce heat loss through glass.

Step 4

Calculate the value of this temperature on the kelvin scale.

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Answer

To convert from Celsius to Kelvin, we use the formula:

K=°C+273.15K = °C + 273.15

Substituting the given temperature:

K=4+273.15=269.15extKK = -4 + 273.15 = 269.15 ext{ K}

Therefore, the temperature on the Kelvin scale is 269.15 K.

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