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## 4.3.3 Thermal properties of materials

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**Objective**(a) define and apply the concept of specific heat capacity**Objective**(b) select and apply the equation E = mcΔθ**Specific Heat Capacity**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**Specific Heat Capacity**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)**Specific Heat Capacity**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)**Specific Heat Capacity**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**Specific Heat Capacity**• How much energy must be supplied to raise the temperature of 5 kg of water from 20°C to 100°C? • 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? • 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**SHC Practical**12 V • Run for 1200 seconds • Take temperature every 30 s • Draw graph • Calculate c • Repeat for different substance heater thermometer metal block**SHC Practical**• calculate gradient Δθ Δt • ΔQ = mc Δθ • VIΔt = mc Δθ • c = VI Δt m Δθ • c = VI 1 m Δt Δθ temp (°C) time (s)**Objective**d) describe what is meant by the terms latent heat of fusion and latent heat of vaporisation**Specific Latent Heat**What is happening at: • AB? • CD? E C 100 D θ • AB: Melting – particles becoming disordered • CD: Boiling – particles completely separating A 0 B O t**Specific Latent Heat**• 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**Specific Latent Heat**• 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**Specific Latent Heat**• 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**Specific Latent Heat**• 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**Definitions**• 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**Specific Latent Heat**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**Specific Latent Heat**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**Specific Latent Heat**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**Specific Latent Heat**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. • 1.0 kg (1000 g) of water requires 2.26 MJ of energy • Therefore, energy = 2.0/1000 x 2.26 x 106 = 4520 J**Specific Latent Heat**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. • 1.0 kg (1000 g) of ice requires 330 kJ of energy • Therefore, energy = 2.0/1000 x 3.30 x 105 = 660 J