4.3.3 Thermal properties of materials

Objective

(a) define and apply the concept of specific heat capacity

Objective

(b) select and apply the equation E = mcΔθ

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If we heat matter so that its temperature rises, the amount of energy we must supply depends on three things:

The mass m of the material The temperature rise Δθ we wish to achieve The material itself

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ΔQ = mcΔθ

where

ΔQ = energy supplied (J)

m = mass (kg)

c = specific heat capacity (J kg-1 K-1)

Δθ = change in temperature (°C or K)

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The specific heat capacity of a substance is numerically equal to the amount of energy required to raise the temperature of 1kg of the substance by 1 K (or by 1 °C)

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When 26 400 J of energy is supplied to a 2kg block of aluminium, its temperature rises from 20 °C to 35 °C. Find the specific heat capacity of aluminium.

c = ΔQ / mΔθ

c = 26 400 J / (2 kg x 15 K)

c = 880 J kg-1 K-1

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a) How much energy must be supplied to raise the temperature of 5 kg of water from 20°C to 100°C?

b) Which requires more energy, heating a 2 kg block of lead by 30 K, or heating a 4 kg block of copper by 5 K?

c) A well-insulated 1 kg block of iron is heated using a 50 W heater for 5 min. Its temperature rises from 22°C to 55°C. Find the specific heat capacity of iron.

Objective

c) describe an electrical experiment to determine the specific heat capacity of a solid or liquid

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metal block

thermometerheater

12 V

Run for 1200 seconds Take temperature every 30 s Draw graph Calculate c Repeat for different substance

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time (s)

tem

p (°

C)

calculate gradient Δθ Δt

ΔQ = mc Δθ

VIΔt = mc Δθ

c = VI Δtm Δθ

c = VI 1m Δt

Δθ

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Objective

d) describe what is meant by the terms latent heat of fusion and latent heat of vaporisation

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O

AB

C

D

E

100

0

θ

t

What is happening at: AB? CD?

AB: Melting – particles becoming disordered

CD: Boiling – particles completely separating

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At AB and CD, energy is being input, but the temperature isn’t rising

The energy is being used to break the molecules free, not raise the temperature

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At AB (melting) and CD (boiling)

energy input temperature does not change molecules become disordered (AB) or separate

from each other (CD) little change in kinetic energy electrical potential energy increases

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At OA and BC and DE

energy input temperature rises molecules move faster kinetic energy increases (temperature = average

ke) little change in electrical potential energy

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The energy needed to cause this change of state is Latent Heat (“Latent” means “hidden”)

When a substance melts, this is the latent heat of fusion

When a substance boils, this is the latent heat of vaporisation

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Latent heat of fusion is the energy which must be supplied to cause a substance to melt at a constant temperature

Latent heat of vaporisation is the energy which must be supplied to cause a substance to boil at a constant temperature

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Remember: Temperature is a measure of the average

kinetic energy of the molecules When a thermometer is put into water, the

water molecules collide with the thermometer and share their kinetic energy with it.

At a change of state, there is no change in kinetic energy, so no change of temperature

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Why does it take more energy to boil a substance than it does to melt it?

Melting – molecules still bonded to most of their neighbours – breaks one or two bonds

Boiling – each molecule breaks free from all of its neighbours – breaks eight or nine bonds

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The specific latent heat of a substance is the energy required per kilogram of the substance to change its state without any change of temperature

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The specific latent heat of vaporisation of water is 2.26 MJ kg-1. Calculate the energy needed to change 2.0 g of water into steam at 100 °C.

1. 1.0 kg (1000 g) of water requires 2.26 MJ of energy

2. Therefore,

energy = 2.0/1000 x 2.26 x 106

= 4520 J

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The specific latent heat of fusion of water is 330 kJ kg-1. Calculate the energy needed to change 2.0 g of ice into water at 0 °C.

1. 1.0 kg (1000 g) of ice requires 330 kJ of energy

2. Therefore,

energy = 2.0/1000 x 3.30 x 105

= 660 J

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