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School-college Physics Notes: Thermal energy 5.5 Density

Density and particle theory: 5.5 Calculations involving density data, density formula - practice questions for objects of different shapes!

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5.6 Calculations involving density

Q1 Irregular shaped solid object

A stone weighing 27.2 g displaced 8.5 cm3 of water (8.5 ml), calculate its density in (a) in g/cm3 and then (b) in kg/m3.

Q2 A regular solid block

A block of iron had dimensions of 3.0 cm x 5.0 cm x 12.0 cm and weighed 1.420 kg, calculate the density of iron.

Q3 A regular solid cylinder of an alloy has a diameter of 3.0 cm, a length of 12.0 cm and a mass of 750 g.

Calculate the density in kg/m3

Tricky, involves several unit conversions

Q4 A cube of material has a side length of 2.5 cm.

If the material has a density of 5000 kg/m3, what is the mass of the cube in g and kg?

Q5 If the density of air is 1.30 kg/m3

(a) Calculate the mass of air in a room measuring 7 m by 6 m by 3 m.

(b) Explain what happens to the density of air in the room if the atmospheric pressure increases.

Q6 Steel has been cast into long rectangular bars 25 metres long and a cross-section of 20 cm by 30 cm.

If the density of steel is 7900 kg/m3, what is the minimum number of bars needed to transport at least 230 tonnes of the steel?

Keywords, phrases and learning objectives for density

Practice exam questions on density problem solving

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Worked out ANSWERS to the density questions

Q1 Irregular shaped solid object

A stone weighing 27.2 g displaced 8.5 cm3 of water (8.5 ml), calculate its density.

(a) density of solid ρ = m ÷ v, ρ = 27.2 ÷ 8.5 = 3.2 g/cm3

(b) However, you may have to calculate the density in kg/m3 and this is arithmetically a bit more awkward!

27.2 g = 27.2/1000 = 0.0272 kg   and   8.5 cm3 = 8.5 / 106 = 8.5 x 10-6 m3    (1 m3 = 106 cm3)

ρ = m ÷ v, ρ = 0.0272 ÷ (8.5 x 10-6) = 0.0272 ÷ 0.0000085 = 3200 kg/m3

It might be handy to know that kg/m3 is 1000 x g/cm3 !!! 3.2 x 1000 = 3200 !!!

Q2 A regular solid block

A block of iron had dimensions of 3.0 cm x 5.0 cm x 12.0 cm and weighed 1.420 kg, calculate the density of iron.

Volume of block = 3 x 5 x 12 = 180 cm3,  volume = 180/106 = 1.8 x 10-4 m3   (0.00018)

density of solid ρ = m ÷ v, ρ = 1.42 ÷ 0.00018 = 7889 kg/m3

Q3 A regular solid cylinder of an alloy has a diameter of 3.0 cm, a length of 12.0 cm and a mass of 750 g.

Calculate the density in kg/m3

radius of cylinder = diameter/2 = 3.0/2 = 1.5 cm,

cross-section area = πr2 = 3.142 x 1.52 = 7.070 cm2

volume of cylinder = 7.070 x 12.0 = 84.83 cm3, 84.83/106 = 8.483 x 10-5 m3   (remember 1 m3 = 106 cm3)

mass of cylinder = 750/1000 = 0.750 kg  (1 kg = 1000 g)

density of solid ρ = m ÷ v, ρ = 0.750 ÷ 8.483 x 10-5 = 8841 = 8840 kg/m3  (3 sf)

Q4 A cube of material has a side length of 2.5 cm.

If the material has a density of 5000 kg/m3, what is the mass of the cube in g and kg?

l = 2.5/100 = 0.025 m, therefore volume = l3 = 0.0253 = 1.5625 x 10-5 m3

ρ = m ÷ v,  m = ρ x v = 5000 x 1.5625 x 10-5 = 0.078 kg (x 1000 = 78 g)

Q5 If the density of air is 1.30 kg/m3

(a) Calculate the mass of air in a room measuring 7 m by 6 m by 3 m.

Volume of room = l x b x h = 7 x 6 x 3 = 126 m3

ρ = m ÷ v,  m = ρ x v = 1.3 x 126 = 164 kg (3 sf)

(b) Explain what happens to the density of air in the room if the atmospheric pressure increases.

If the pressure increases, on average, the particles are squashed closer together.

Therefore, there is more mass in the same volume, so the density increases.

Note: (i) The mass of air in the room will also increase.

(ii) In any situation where a gas is compressed, the density of the gas is increased.

Q6 Steel has been cast into long rectangular bars 25 metres long and a cross-section of 20 cm by 30 cm.

If the density of steel is 7900 kg/m3, what is the minimum number of bars needed to transport at least 230 tonnes of the steel?

V = l x b x h, l = 25 m, b = 20/100 = 0.2 m, h = 30/100 = 0.3 m

V of 1 rod = 1.5 m3

ρ = m ÷ v,  m = ρ x v

mass of 1 bar = 7900 x 1.5 = 11850 kg.

230 tonnes ≡ 230 000 kg (1 metric tonne = 1000 kg)

bars needed = 230000/11850 = 19.4 bars.

So you would need 20 bars to transport a minimum of 230 tonnes of steel.

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