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GCSE · Physics

IG.P10 Density, Pressure in Liquids and Change of State

EDEXCEL 4PH1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Define density as it is used in the context of solids and liquids, giving the equation and naming each symbol.

(Total for Question 1 is 1 mark)

State the density of water used for simple comparisons at standard conditions, and give its units.

(Total for Question 2 is 1 mark)

A small metal cube used in a school test has mass 0.48 kg and volume 60 cm³. Calculate its density in kg/m³.

(Total for Question 3 is 1 mark)

An object floats on oil but sinks in water. Explain, using density ideas, what this tells you about the density of the object compared with oil and water.

(Total for Question 4 is 2 marks)

Define pressure in a liquid at a depth and state the equation linking pressure, depth, fluid density and gravitational field strength, naming symbols.

(Total for Question 5 is 2 marks)

A diver swims at a depth of 12 m in seawater with density 1025 kg/m³. Calculate the increase in pressure due to the water at that depth. Use g = 9.8 N/kg. Give your answer in Pa.

(Total for Question 6 is 3 marks)

Explain why the pressure at the base of a tall tank of mercury is greater than at the base of a tall tank of water of the same depth, using the p = h ρ g equation and particle ideas.

(Total for Question 7 is 3 marks)

A sealed container has two immiscible liquid layers of equal depth 0.20 m. The lower layer is oil with density 850 kg/m³ and the upper layer is water with density 1000 kg/m³. Calculate the pressure at the interface between the oil and water due to the oil alone. Use g = 9.8 N/kg. Give your answer in Pa.

(Total for Question 8 is 2 marks)

State a typical value for atmospheric pressure at sea level and give its units.

(Total for Question 9 is 1 mark)

A block of aluminium of mass 270 g is heated, and its temperature rises from 20 degrees C to 80 degrees C. The specific heat capacity of aluminium is 900 J/kg degrees C. Calculate the energy required to heat the block. Show equation, substitution and final answer with units.

(Total for Question 10 is 3 marks)

A student uses 500 g of water at 20 degrees C and supplies 9,000 J of energy. Calculate the final temperature of the water. Use specific heat capacity of water as 4200 J/kg degrees C. Show equation, substitution and final answer with units.

(Total for Question 11 is 4 marks)

Ice of mass 0.25 kg at 0 degrees C is melted to water at 0 degrees C. The specific latent heat of fusion of ice is 3.34 x 10⁵ J/kg. Calculate the energy required. Show equation, substitution and final answer with units.

(Total for Question 12 is 3 marks)

A kettle contains 0.35 kg of water at 100 degrees C and some of the water is converted to steam at 100 degrees C. If 4.6 x 10⁵ J of energy is supplied to the kettle to boil water into steam, calculate the mass of water that becomes steam. Use specific latent heat of vaporisation L = 2.26 x 10⁶ J/kg. Show equation, substitution and final answer with units.

(Total for Question 13 is 3 marks)

Explain, using particle ideas, why the temperature of a substance stays constant during melting or boiling even while energy is still being supplied. Give four distinct points.

(Total for Question 14 is 4 marks)

A small sample of an unknown metal has volume 15.0 cm³ and mass 120 g. Calculate its density in kg/m³ and suggest which metal it could be from this list: aluminium (2700 kg/m³), copper (8900 kg/m³), iron (7800 kg/m³), zinc (7100 kg/m³). Show your working and final answer.

(Total for Question 15 is 3 marks)

A cube of plastic has edge length 6.0 cm and density 950 kg/m³. Calculate the mass of the cube in grams. Show equation, substitution and final answer with units.

(Total for Question 16 is 2 marks)

Using p = h ρ g, calculate how much the pressure increases per metre depth in freshwater (ρ = 1000 kg/m³). Give your answer in Pa per metre. Use g = 9.8 N/kg.

(Total for Question 17 is 2 marks)

Mark scheme · IG.P10 Density, Pressure in Liquids and Change of State

Question 1

B1 ρ = mass / volume, with ρ = density, mass = mass (kg) and volume = volume (m³) or equivalent units
Answer: ρ = mass / volume, where ρ is density, mass is mass and volume is volume

Question 2

B1 1000 kg/m³
Answer: 1000 kg/m³

Question 3

M1 convert 60 cm³ to m³ = 60 x 10^-6 m³ or 6.0 x 10^-5 m³
A1 density = 0.48 / 6.0 x 10^-5 = 8000 kg/m³ cao
Answer: 8000 kg/m³

Question 4

B1 floats on oil means object density < density of oil
B1 sinks in water means object density > density of water
Answer: Object density is less than oil density but greater than water density

Question 5

B1 p = h x ρ x g with p = pressure, h = depth, ρ = fluid density, g = gravitational field strength
B1 states that pressure increases with depth and with fluid density
Answer: p = h ρ g, where p is pressure, h depth, ρ fluid density and g gravitational field strength; pressure increases with depth and density

Question 6

M1 writes p = h ρ g
M1 substitutes p = 12 x 1025 x 9.8
A1 p = 120,540 Pa (give or take rounding) cao
Answer: 1.21 x 10⁵ Pa (120540 Pa)

Question 7

B1 states p = h ρ g and that h and g are same so difference is ρ
B1 mercury has higher density (more mass per unit volume) than water
B1 therefore weight of column above base is greater, so pressure at base is greater
Answer: Because mercury has a larger density, the term ρ is bigger so p = h ρ g is larger; a denser fluid means more mass in the same column and therefore more weight on the base

Question 8

M1 uses p = h ρ g with h = 0.20, ρ = 850, g = 9.8
A1 p = 0.20 x 850 x 9.8 = 1666 Pa (rounding acceptable) cao
Answer: Approximately 1.67 x 10³ Pa (1666 Pa)

Question 9

B1 1.01 x 10⁵ Pa or 101,325 Pa
Answer: 1.01 x 10⁵ Pa (approximately 101,325 Pa)

Question 10

M1 writes E = m c DeltaT
M1 substitutes m = 0.270 kg, c = 900, DeltaT = 60 so E = 0.270 x 900 x 60
A1 E = 14,580 J cao with units
Answer: E = 1.46 x 10⁴ J (14,580 J)

Question 11

M1 writes E = m c DeltaT
M1 substitutes E = 9000, m = 0.500 kg, c = 4200 so 9000 = 0.500 x 4200 x DeltaT
M1 rearranges to DeltaT = 9000 / (0.500 x 4200)
A1 DeltaT = 4.29 degrees C so final temperature = 24.29 degrees C cao
Answer: Final temperature = about 24.3 degrees C

Question 12

M1 writes E = m L
M1 substitutes m = 0.25, L = 3.34 x 10⁵ so E = 0.25 x 3.34 x 10⁵
A1 E = 8.35 x 10⁴ J cao
Answer: E = 8.35 x 10⁴ J (83,500 J)

Question 13

M1 writes E = m L and rearranges to m = E / L
M1 substitutes m = 4.6 x 10⁵ / 2.26 x 10⁶
A1 m = 0.2035 kg (allow awrt 0.204 kg) cao
Answer: Approximately 0.204 kg of water becomes steam

Question 14

B1 states that supplied energy is used to break bonds or overcome attractive forces between particles rather than increase kinetic energy
B1 states that average kinetic energy of particles (and so temperature) stays the same while bonds are being broken
B1 mentions change of arrangement or spacing of particles, e.g. particles move further apart when melting or boiling
B1 links these points to latent heat, energy required per kg for the change of state
Answer: Energy goes into breaking intermolecular forces and increasing particle separation rather than raising particle kinetic energy, so temperature stays constant during the change of state; this energy per kg is the latent heat

Question 15

M1 converts mass to kg and volume to m³: m = 0.120 kg, V = 15.0 x 10^-6 m³
M1 computes density = m / V = 0.120 / 15.0e-6 = 8000 kg/m³
A1 identifies closest metal as iron (7800 kg/m³) or states value 8000 kg/m³ and suggests iron as reasonable match
Answer: Density = 8.00 x 10³ kg/m³, suggests iron as the closest match

Question 16

M1 calculates volume = 6.0³ cm³ = 216 cm³ = 216 x 10^-6 m³
A1 mass = density x volume = 950 x 216e-6 = 0.2052 kg = 205.2 g cao
Answer: Mass = 205.2 g

Question 17

M1 uses p/h = ρ g = 1000 x 9.8
A1 pressure increases by 9,800 Pa per metre cao
Answer: Pressure increases by 9,800 Pa per metre

Density, Pressure in Liquids and Change of State - Worksheets, Questions and Revision