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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. 1 2 (b) Figure 3 shows the dimensions of a solid block ... 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, the particles are closely packed together and vibrate in fixed positions. In contrast, the particles in a liquid are also close together but are able to move freely around each other.

  2. The particles in a solid have a regular arrangement, while the particles in a liquid are randomly arranged. Additionally, particles in a liquid have more kinetic energy compared to those in a solid.

Step 2

Calculate the mass of the concrete block.

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Answer

To calculate the mass of the concrete block, we first need to find its volume. The dimensions given are:

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

The formula for volume (V) is:

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

So,

V=1.5imes1.0imes0.20=0.30extm3V = 1.5 imes 1.0 imes 0.20 = 0.30 ext{ m}^3

Now, we can use the density formula to find the mass (m):

ho imes V$$ Where: - Density, $\rho = 2100 \, \text{kg/m}^3$ - Volume, $V = 0.30 \, \text{m}^3$ Thus, $$m = 2100 imes 0.30 = 630 \, \text{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 of the shed to minimize heat transfer.

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

Step 4

Calculate the value of this temperature on the kelvin scale.

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Answer

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

K=°C+273.15K = °C + 273.15

For -4°C:

K=4+273.15=269.15KK = -4 + 273.15 = 269.15 \, K

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