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

13 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 4)Density, Pressure in Liquids and Change of State revision guide

This topic is chapter 10 of IGCSE Physics Practice Book.

3.10 Density, Pressure in Liquids and Change of State

EDEXCEL 4PH1 · Calculator allowed · about 45 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 a typical value for atmospheric pressure at sea level 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 2 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 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)

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 6 is 3 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 7 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 8 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 9 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 10 is 2 marks)

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 11 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 12 is 4 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)

Mark scheme · 3.10 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 1.01 x 10⁵ Pa or 101,325 Pa
Answer: 1.01 x 10⁵ Pa (approximately 101,325 Pa)

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

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 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 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 7

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

Question 8

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 9

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 10

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 11

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 12

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 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

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

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3.10 Density, Pressure in Liquids and Change of State

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