# FXA Candidates should be able to : Define and apply the concept of specific heat capacity. Select and apply the equation : E = mcδθ

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1 UNIT G484 Module Thermal Properties of Materials 1 Candidates should be able to : Define and apply the concept of specific heat capacity. Select and apply the equation : E = mcδθ The MASS (m) of the object. The SMALLER the mass of the object, the LARGER the temperature rise (or fall) for the same heat supplied to (or extracted from) the object. Describe an electrical experiment to determine the specific heat capacity of a solid or a liquid. Δθ α 1/m Describe what is meant by latent heat of fusion and latent heat of vaporisation. The MATERIAL of the object. SPECIFIC HEAT CAPACITY Objects of the same mass, but different material, have a different temperature rise (or fall) with the same amount of heat supplied to (or extracted from) them. This fact is taken into account by a quantity called the HEAT CAPACITY of the object. The TEMPERATURE RISE (or FALL) (Δθ) of an object depends on : The HEAT ENERGY (E) supplied to (or extracted from) the object. The HEAT CAPACITY of an object is the amount of heat energy which must be supplied to (or extracted from) it, to make its temperature rise (or fall) by 1 K (or 1 C). The GREATER the amount of heat supplied to (or extracted from) an object, the GREATER is the temperature rise (or fall) for a given mass. Δθ α E So the GREATER the heat capacity of the object, the SMALLER the temperature rise (or fall) for the same amount of heat supplied to (or extracted from) it. HEAT CAPACITY only relates to a particular object. A much more useful quantity which is a property of a material, is SPECIFIC HEAT CAPACITY (c).

2 UNIT G484 Module Thermal Properties of Materials 2 The SPECIFIC HEAT CAPACITY (c) of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1 K (1 C) without a change of state. MEASUREMENT OF SPECIFIC HEAT CAPACITY OF A SOLID low voltage d.c. supply The unit of (c) is : J kg -1 K -1 or J kg -1 C -1 The table below gives some typical SPECIFIC HEAT CAPACITY values : SUBSTANCE c/j kg -1 K -1 Aluminium 900 Concrete 850 Copper 390 thermometer insulation immersion heater metal block Iron 490 Lead 130 The mass (m) of the metal block is accurately measured using an electronic top-pan balance. Oil 2100 Water 4200 The variable resistor is adjusted so as to give suitable values of voltage (V 1 ) and current (I 1 ). The supply is then switched off and the initial temperature (θ i ) of the block is measured and recorded. The amount of ENERGY (E) which must be supplied to a MASS (m) of a substance of SPECIFIC HEATCAPACITY (c) in order to produce a TEMPERATURE RISE (Δθ) is given by : The supply is then switched on and the block is heated for a fixed time (say 10 minutes). The final temperature (θ f ) of the block is also measured and recorded. Then, temperature rise of the block, Δθ 1 = θ f - θ i E = m c Δθ (J) (kg) (J kg -1 K -1 ) (K) After allowing the block to cool down to room temperature, the procedure is repeated using lower voltage (V 2 ) and current (I 1 ) values, as adjusted with the variable resistor. The block is heated for the same time as before and the new temperature rise, Δθ 1 is calculated as before.

3 UNIT G484 Module Thermal Properties of Materials PRACTICE QUESTIONS 3 RESULTS Mass of block, m = Time for which block is heated, t = kg s 1 Using the table of specific heat capacities given on page 2, calculate the energy which must be supplied (or extracted) to raise (or lower) the temperature of : Voltage across heater V 1 /V V 2 /V Current in heater I 1 /A I 2 /A (a) An iron ingot of mass 12 kg from 20 C to 190 C. (b) 250 g of water from 18 C to 100 C. (c) A copper ring of mass 45 g from 20 C to -140 C. Initial temperature of block θ i / C θ i / C Final temperature of block θ f / C θ f / C Temperature rise of block Δθ 1 / C Δθ 2 / C CALCULATIONS 2 Calculate the time taken to raise the temperature of 300 g of oil in an aluminium, deep-fat frier of mass 0.75 kg from 20 C to 145 C, if the electric heating element has a power rating of 3.0 kw. Why is your calculated time an under-estimate? electrical energy supplied = heat energy gained + heat energy lost to to the block. by the block. the surroundings. 3 A 3kW electric kettle took 260 s to heat 1.8 kg of water from 18 C to 100 C. V 1 I 1 t = m c Δθ 1 + H.. (1) V 2 I 2 t = m c Δθ 2 + H.. (2) (1) - (2) : V 1 I 1 t - V 2 I 2 t = m c (Δθ 1 - Δθ 2 ) (a) Calculate : (i) The electrical energy supplied to the kettle in this time. (ii) The internal energy gained by the water in the kettle. From which : c = V 1 I 1 t - V 2 I 2 t m (Δθ 1 - Δθ 2 ) c = c = J kg -1 K -1 (b) The mass of the kettle was 1.2 kg and it was made from aluminium. (i) Calculate the internal energy gained by the kettle. (ii) Account for the difference between the electrical energy supplied to the kettle and the internal energy gained by the water and kettle. Use the table on page 2 to obtain the values you need for the specific heat capacities of water and aluminium.

4 UNIT G484 Module Thermal Properties of Materials 4 LATENT HEAT LATENT HEAT is the heat energy which a body will absorb during MELTING or EVAPORATION and which it gives out during SOLIDIFICATION (FREEZING) or CONDENSATION. It is called latent (meaning hidden) because it produces a CHANGE OF STATE, but NO TEMPERATURE CHANGE. Starting from the left, with a solid well below its MELTING POINT, we see that the heat supplied causes the temperature of the solid to increase until it reaches the MELTING POINT (θ m ). Continued heating does not cause any further increase in temperature until ALL the solid has melted. Continuous extraction of heat energy at a constant Continuous supply of heat energy at a constant rate The heat supplied (or extracted) to cause the phase change from SOLID TO LIQUID (or LIQUID to SOLID) without a change of temperature, is called the LATENT HEAT OF FUSION. Temperature/ C Boiling or condensing Melting or solidifying When all the solid has melted, the temperature of the resulting liquid then increases until the BOILING POINT (θ b ) is reached and VAPORISATION occurs throughout the volume of the liquid (and not just at the surface). θ b Continued heating does not cause any further increase in temperature until until ALL the liquid has vaporised. θ m 0 SOLID SOLID + LIQUID LIQUID LIQUID + GAS GAS Time/s The heat supplied (or extracted) to cause the phase change from LIQUID TO VAPOUR (or VAPOUR TO LIQUID) Without a change of temperature, is called the LATENT HEAT OF VAPORISATION. θ b is the Boiling point. θ m is the melting point. The diagram shown above illustrates what happens when a SOLID is continuously supplied with heat energy, or when heat energy is continuously extracted from a GAS. If the diagram is read from left to right, it shows the sequence of events When heat energy is continuously added to a solid, while reading it from right to left Illustrates the reverse process of extracting heat energy from a gas. Once all the liquid has vaporised, continued heating of the gas or vapour will again cause the temperature to rise.

5 UNIT G484 Module Thermal Properties of Materials HOMEWORK QUESTIONS 5 1 (a) Define specific heat capacity. The SPECIFIC LATENT HEAT (L) OF FUSION (or VAPORISATION) of a substance is the energy needed to change 1 kg of the substance from SOLID TO LIQUID (or LIQUID TO VAPOUR) at constant temperature. (b) Describe an electrical method for determining the specific heat capacity (c) of copper. Give a labelled diagram of the apparatus used and show how the results of such an experiment are used to calculate a value for (c). The ENERGY (E) which must be supplied (or extracted) to cause a change of state in a MASS (m) of a substance having a SPECIFIC LATENT HEAT (L) is given by : E = ml (J) (kg) (J kg -1 ) (c) In an experiment of the type you have described in (b), a 1.5 kg copper block was heated for 15 minutes using an electric immersion heater which is embedded in it. The temperature of the block was found to rise by 36 C when the p.d. across the heater was 12.0 V and the current in it was 2.0 A. When the experiment was repeated with p.d. and current values of 10.0 V and 1.9 A respectively and the same heating time, the new observed temperature rise was 28.5 C. EVAPORATION AND BOILING Use the two sets of results to calculate the specific heat capacity of copper. EVAPORATION is the name given to the process by which a liquid changes state into a gas or vapour. Evaporation occurs : AT ANY TEMPERATURE. ONLY FROM THE LIQUID SURFACE. 2 A convector heater having a power rating of 2.5 kw is used to heat an office whose dimensions are 6.0 m x 5.0 m x 2.8 m. If the initial temperature of the air in the office is 12 C, calculate the time taken for the heater to raise the air temperature to a comfortable 20 C. BOILING is a special case of evaporation. Boiling occurs : AT A SPECIFIC TEMPERATURE (THE BOILING POINT). THROUGHOUT THE LIQUID VOLUME. Bubbles of vapour form inside the liquid, rise to the surface and burst, releasing the vapour to the atmosphere. (Density of air at standard atmospheric pressure = 1.2 kg m -3 ). Why will your calculated time be less than the time it will actually take to produce the required temperature change?

6 UNIT G484 Module Thermal Properties of Materials 3 An insulated metal cylinder of mass 1.4 kg was heated by a 24 W electric heater placed in a slot in the cylinder. The temperature of the metal was measured using a thermometer placed in a different slot in the metal. The measurements obtained in the experiment were used to plat a graph of temperature against time as shown in the diagram below. Temperature/ C 40 4 (a) Draw a well-labelled graph of temperature against time to 6 show what happens when a solid object is continuously supplied with heat energy until some time after it has become a gas. Indicate those sections of the graph where the object is solid, solid + liquid, liquid, liquid + gas and gas. Show also the positions of the melting and boiling points on the temperature axis. (b) Define latent heat of fusion and latent heat of vaporisation. (c) Explain the difference between evaporation and boiling Time/s (a) Determine the temperature rise per second of the cylinder. (b) Calculate the specific heat capacity of the metal of which the cylinder is made.

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