EJC Physics H209 First Law of Thermodynamics 2024 1. Notes (FULL)

Aircon Compressor s. Air-conditioned environments provide a welcome respite from the tropical humidity . “Blower” units in interiors have refrigerant connected to the compressor which usually sits outdoors. The compressor pressurizes the refrigerant and passes the heated liquid through fins cooled down by a fan. The refrigerant is further cooled by adiabatic expansion, and re-circulated back into the blower units, ready to transport more thermal energy from the interiors. As a cyclic process, there is net work done on the refrigerant i.e. there is more energy usage to operate the aircon system than the thermal energy the aircon system is able to extract. Content • Specific heat capacity and specific latent heat • Internal energy • First law of thermodynamics Learning Outcomes Candidates should be able to: (a) define and use the concepts of specific heat capacity and specific latent heat (b) show an understanding that internal energy is determined by the state of the system and that it can be expressed as the sum of a random distribution of kinetic and potential energies associated with the molecules of a system (c) relate a rise in temperature of a body to an increase in its internal energy (d) recall and use the first law of thermodynamics expressed in terms of the increase in internal energy, the heat supplied to the system and the work done on the system. For the first part of this topic, we return to the macroscopic scale to discuss how heat and temperature relates to work and energy. Devices which convert thermal energy into mechanical motion are typically referred to as engines. We then bridge the above to the microscopic view and see how the interplay between energy and useful work is explained by the kinetic model of gases. 4-stroke internal -combustion petrol engine. The air -fuel mixture is compressed almost instantaneously – i.e. work is done on the gas. A spark ignites the air-fuel mix, resulting in great increases in pressure. The piston is pushed downwards and the gas does work by exerting the force (pressure applied on the piston surface) downwards and turning the crank shaft.

Similar to the idea of “GDP per capita” (normalizes economic output to per person basis), “population density” (normalizes number of people to per area basis) and “density” (normalizes mass to per unit volume basis), “specific” in thermal physics normalizes the quantity to a per unit mass basis. A specific quantity in thermal physics allows us to discuss an intrinsic property of a material. Since specific heat capacity discusses the property of a material, notice the definition involves substance rather than a body. Experiments for finding specific heat capacity need to a find the mass, b measure the temperature rise, c determine the quantity of thermal energy supplied (often by electrical means so E = Pt = Q) d(i) prevent unwanted heat loss from the setup to surrounding OR d(ii) permit the heat loss at a desired rate so as to accommodate calculations involving it. Example 1 A 2.5 kg block of copper is wrapped in insulation. The block has 2 holes drilled to fit a thermometer and an electric heater. The heater has a constant effective resistance of 9.6 Ω and is connected to a benchtop constant-voltage supply. A rise of 50 C is seen after 36.0 V is supplied for 6 minutes. Determine the specific heat capacity of the copper material. By Principle of Conservation of Energy ( ) ( ) ( )( )( ) 2 22 11 energy supplied energy to by heater raise temperature 36 0 6 60 9 6 2 5 50 389 J kg K Pt m T V t mc TR .Vtc R T . . c m −−  = ==  = = = Note: Assumed all thermal energy by heater transferred to copper block without loss to surroundings. Specific heat capacity is the thermal energy per unit mass to raise the temperature of a substance by one degree. ( )recall Qm Tc T Qc m = =  [S.I. unit: J kg-1 K-1] c = specific heat capacity Q = quantity of thermal energy ΔT = change in temperature tabletop lagging solid copper thermometer constant voltage power supply A V heater

Example 2 A 250 W heater is used to heat up 300 g of sea water contained in a calorimeter with a heat capacity of 230 J K-1. The temperature of the sea water was initially 25.0 C. After 130 s of heating, it reached a maximum of 45.0 C. Find the specific heat capacity of sea water. State an assumption you made in your calculation. By Principle of Conservation of Energy ( )( ) ( ) ( )( ) − − − =+ =  +  −−==  − = 3 1 energy to energy toenergy supplied raise temperature raise temperatur e by heater of sea water of calorimeter 250 130 230 45 0 25 0 300 10 45 0 25 0 4650 J kg K T C T ..Pt C Tc mT . Pt mc . −1 Assumed all thermal energy from heater transferred to sea water and calorimeter without loss to surroundings. Example 3 (a) to account for heat loss from tube to surrounding [not allowed: stop/prevent heat loss] (b) By Principle of Conservation of Energy, energy from energy to raise heat loss to heater temperature surrounding=+ ( ) ( ) ( )( )( ) loss 1 1 loss 2 2 loss 1 2 1 2 12 12 3 11 in 1 second, ; 1 58 1 11 10 25 5 19 5 4110 J k 44 9 33 3 g K P m c P m P mc T h P m c T h P m c T h P m T Pc . . . .. mT .− −− =  + =  + =  + − −  −= − = −  − = − = Liquid enters a tube at a constant temperature of 19.5 °C. The mass of liquid flowing through the tube per unit time is m. Electrical power P is dissipated in the heating coil. A student found 2 sets of m and P where liquid exits the tube at 25.5 °C. (a) Suggest why the student obtains two sets of data rather than one. (b) Find the specific heat capacity of the liquid. calorimeter stirrer thermometer heater m / g s-1 P / W 1.11 33.3 1.58 44.9 Note: Recall that temperature difference provides 2 pieces of info, (i) direction of heat flow (from heated water to surrounding) and (ii) rate of heat transfer (the larger the difference in temperature, the faster the rate). This continuous flow method is suitable for gases and liquids. The temperature at the inlet and outlet is the same across 2 data sets (so temperature difference between inlet vs surrounding is same; temperature difference between outlet vs surrounding is another but same value) – so rate of heat loss to surrounding, hloss, is same for both experiment runs.

Example 4 (a) t = 0 because there is no temperature difference with surrounding. (b) (Using gradient near t = 0, extend straight line to T = 84 C.) (c) By Principle of Conservation of Energy in the case of no heat loss, energy from energy to raise heat loss to heater temperature surrounding=+ ( )( ) ( )( ) 3 11 95 6 67 60 10 84 24 85 0 0 J kg K Pt mc Pt m T c T − −−  =  = − = = An aluminum heatsink of mass 670 g is initially at room temperature of 24 C. It is simultaneously heated at constant rate of 95 W and cooled by a fan. The variation with time t of the temperature T is shown. (a) Suggest the time at which heat loss to surrounding is minimum. (b) Hence, using the Figure, determine the final temperature if there were no heat loss to surrounding. (c) Find the specific heat capacity of aluminium. T cooling fan heatsink

A pure substance will undergo phase change at one constant temperature, whereas mixtures will under phase change over a range of temperatures. A phase change takes time – during the transition the phases coexist (across a phase boundary) until the phase change is completed. *To adapt the definition for specific latent heat of fusion, add the phrase “ from solid phase to liquid phase” after “change the phase of a substance…” Similarly, for specific latent heat of vaporisation, add the phrase “from liquid phase to gas phase” after “change the phase of a substance…” Experiments for finding specific latent heat capacity need to a find the mass,