TJC 2023 IP4 Themal Properties of Matter Notes

IP4 Physics Topic 19 – Thermal Properties of Matter Page 1 of 9 TJC IP4 PHYSICS Student’s Copy Topic 19 – Thermal Properties of Matter Content • Internal energy • Specific heat capacity • Melting, boiling and evaporation • Specific latent heat Learning Outcomes Candidates should be able to: (a) describe a rise in temperature of a body in terms of an increase in its internal energy (random thermal energy) (b) define the terms heat capacity and specific heat capacity (c) recall and apply the relationship thermal energy = mass x specific heat capacity x change in temperature to new situations or to solve related problems (d) describe melting/solidification and bo iling/condensation as processes of energy transfer without a change in temperature (e) explain the difference between boiling and evaporation (f) define the terms latent heat and specific latent heat (g) recall and apply the relationship thermal energy = mass x specific latent heat to new situations or to solve related problems (h) explain latent heat in terms of molecular behaviour (i) sketch and interpret a cooling curve Note that the following Learning outcomes of Transfer of Thermal Energy had been completed in IP2 Green Science (refer to Physics Matters reference book Chapter 10) 1. show understanding that thermal energy is transferred from a region of higher temperature to a region of lower temperature 2. explain what is meant by conduction, convection and radiation 3. describe, in molecular terms, how energy transfer occurs in solids 4. describe, in terms of density changes, convection in fluids 5. explain that energy transfer of a body by radiation does not require a material medium and the rate of energy transfer is affected by: • colour and texture of the surface • surface temperature • surface area 6. infer from experiments that different materials have different rates of heat flow 7. apply the concept of thermal energy transfer to everyday applications This study source was downloaded by 100000857398866 from CourseHero.com on 11-15-2024 00:08:48 GMT -06:00 https://www.coursehero.com/file/237296757/202320IP420Physics20Topic1920Themal20Properties20of20Matter20Notes20S-2pdf/

IP4 Physics Topic 19 – Thermal Properties of Matter Page 2 of 9 TJC IP4 PHYSICS Student’s Copy 19.1 Internal energy LO (a) All particles in matter vibrate about fixed positions (kinetic energy) and are held together by strong intermolecular forces (potential energy). The total energy for the particles is called the internal energy. The internal energy in molecules consists of: ▪ kinetic energy that directly depends on temperature ▪ potential energy that depends on the force between the molecules and their distance apart ▪ when the temperature of a substance rises, internal energy increases ▪ when the temperature of a substance falls, energy is released and internal energy decreases forces This study source was downloaded by 100000857398866 from CourseHero.com on 11-15-2024 00:08:48 GMT -06:00 https://www.coursehero.com/file/237296757/202320IP420Physics20Topic1920Themal20Properties20of20Matter20Notes20S-2pdf/

IP4 Physics Topic 19 – Thermal Properties of Matter Page 3 of 9 TJC IP4 PHYSICS Student’s Copy 19.2 Thermal Energy in relation to change in temperature LO (b) & (c) Heat Capacity C of an object The quantity/amount of thermal energy or heat absorbed / emitted , Q, by the b ody per unit temperature change (1K or 1°C). ▪ SI unit is J K-1 or J °C-1 ▪ different substances have different heat capacities Specific heat capacity c of an object The quantity/amount of thermal energy or heat absorbed / emitted, Q, per unit mass (1 kg) of the material per unit temperature change (1K or 1°C). ▪ SI unit is J kg-1 K-1 or J kg-1 °C-1 ▪ substances with a high specific heat capacity warm up (or cool) more slowly than substances with a lower heat capacity because they must absorb (or lose) more heat to raise (or lower) the temperature ▪ Examples: specific heat capacity for – water : 4200 J kg-1°C-1 - copper : 400 J kg-1°C-1 Effects and applications of the high specific heat capacity of water Water has a high specific heat capacity compared to other substances. ▪ water needs a lot of energy to warm it up; once warm, it holds a good store of energy ▪ loss of a large amount of energy causes a small drop in temperature ▪ sea temperature rises and falls very slowly The high specific heat capacity of water (as well as its relative cheapness and availability ) accounts for its use ▪ as the circulating liquid in central heating systems ▪ as a cooling liquid in car engines ▪ as hot water bottles to keep people or things warm Worked Example 1 An electric heater of power 800 W raises the temperature of 4.0 kg of a liquid from 30°C to 50°C in 100 s. Calculate (a) the heat capacity of the 4.0 kg liquid, (b) the specific heat capacity of the liquid. (a) Assuming that heat loss to surrounding/ environment is neglected, by conservation of energy, Heat supplied by electric heater = Heat absorbed by container + Heat absorbed by water P t = C T (800 W) (100 s) = C (50 – 30) C = 4000 J°C-1 (b) C = mc → 4000 J°C-1 = (4.0 kg) c → c = 1000 Jkg-1 °C-1 c = Q m∆T m kg Q joules of energy temperature changes by ∆T °C Where Q - thermal energy absorbed ∆T - change in temperature C = ∆T Q This study source was downloaded by 100000857398866 from CourseHero.com on 11-15-2024 00:08:48 GMT -06:00 https://www.coursehero.com/file/237296757/202320IP420Physics20Topic1920Themal20Properties20of20Matter20Notes20S-2pdf/

IP4 Physics Topic 19 – Thermal Properties of Matter Page 4 of 9 TJC IP4 PHYSICS Student’s Copy Worked Example 2 A copper container of mass 350 g and specific heat capacity 0.4 Jg -1°C-1 contains 500g of water (specific heat capacity 4.2 Jg-1°C-1). The water and container are heated from 30 °C to 100 °C by an electrical coil. (a) How much heat is absorbed by the container? (b) How much heat is absorbed by the water? (c) If the heating takes 3 minutes, how many joules of heat doe s the coil supply per second? (Neglect heat loss) (a) Heat absorbed by container = mcT = (350 g) (0.4 Jg-1°C-1) (100 – 30) = 9800 J (b) Heat absorbed by water = mcT = (500 g) (4.2 Jg-1°C-1) (100 – 30) = 147000 J (c) Assuming that heat loss to surrounding/ environment is neglected , by conservation of energy, Heat supplied by electrical coil = Heat absorbed by container + Heat absorbed by water P t = 9800 + 147000 P (3 x 60 s) = 156800 J P = 871 W Practice Question 1 An experiment is carried out to determine the specific heat capacity of an unknown metal, using a 1 kg block of the metal as shown in Fig. 1. The heater is switched on for 500 s. The following readings are obtained: Change in thermometer reading: 50oC Ammeter reading: 5.0 A Voltmeter reading: 8.0 V (a) Calculate the specific heat capacity of the unknown metal. (b) State why it is not advisable to take the thermometer reading immediately after switching off the current. V A d.c. supply This study source was downloaded by 100000857398866 from CourseHero.com on 11-15-2024 00:08:48 GMT -06:00 https://www.coursehero.com/file/237296757/202320IP420Physics20Topic1920Themal20Properties20of20Matter20Notes20S-2pdf/

IP4 Physics Topic 19 – Thermal Properties of Matter Page 5 of 9 TJC IP4 PHYSICS Student’s Copy 19.3 Thermal Ene