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HEAT FLOW & THE FIRST LAW HEAT is the FLOW of ENERGY because of a temperature difference. The units of HEAT are Joules/s or BTU/s or Calories/s BTU (British.

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Presentation on theme: "HEAT FLOW & THE FIRST LAW HEAT is the FLOW of ENERGY because of a temperature difference. The units of HEAT are Joules/s or BTU/s or Calories/s BTU (British."— Presentation transcript:

1 HEAT FLOW & THE FIRST LAW HEAT is the FLOW of ENERGY because of a temperature difference. The units of HEAT are Joules/s or BTU/s or Calories/s BTU (British Thermal Unit) =1055 J =252.02 cal Calorie (c ) = 4.1855 J Food calorie (C) = 1 kcal =1000 cal (c)

2 HEAT FLOW & THE FIRST LAW The Calorie is defined as the amount of heat added 1 gram of water between 14.5 and 15.5 o C to raise it 1 o C. A food calorie or kilocalorie is the amount of heat added to 1 kg of water (a liter) between 16.5 and 17.5 o C to raise it one degree Celsius.

3 HEAT FLOW & THE FIRST LAW The BTU is the amount of heat added to 1 lb of water (averaged over the temperature range of 32 o F to 212 o F) one degree Fahrenheit. Both the BTU and calorie are non- metric units. The calorie will eventually be phased out in preference to the Joule.

4 HEAT FLOW & THE FIRST LAW The heat flow must be made so that thermal equilibrium occurs at each stage of the process and an equation of state may be used. These changes must be done slowly (compared to propagation of molecules, the speed of sound) V = 330 m/s in air at STP.

5 HEAT FLOW & THE FIRST LAW The processes of heating must be reversible. (So the process can traverse the same path on a P-V diagram) There are irreversible processes which are not controlled and the system is not in thermal equilibrium, so these are not considered. These are not considered.

6 HEAT FLOW & THE FIRST LAW In order to measure the heat flow through a substance one needs to know the HEAT CAPACITY C, so one writes: dQ = C dT where dQ is the small heat flow and dT is the small amount of temperature change.

7 HEAT FLOW & THE FIRST LAW For finite changes one writes: ΔQ = C ΔT One can define specific heat capacity c = ΔQ/mΔT for a one gram mass of the substance. Molar heat capacity c = ΔQ/nΔT for one mole of the substance.

8 HEAT FLOW & THE FIRST LAW Where C = m c = n c m = mass and n = number of moles C has units of cal/K c has units of cal/(g K) c has units of cal/mol K Also c = Mc where M is molecular wt (g) (See Table 18-1 for values.)

9 HEAT FLOW & THE FIRST LAW The heat capacity is C = C(V,T) So C depends on how P,V,and T change dQ = C v dT (isochoric) dQ = C p dT (isobaric) (later well see that C p C v ) The measurement of heat capacities is Adiabatic Calorimetry.

10 HEAT FLOW & THE FIRST LAW Heat can also occur during phase changes because changes in molecular form either release or absorb energy. This occurs during melting or freezing or boiling or condensation or during sublimation or deposition. During these processes, the temperature does not change so they are called Latent heats.

11 HEAT FLOW & THE FIRST LAW The Latent Heats are measured in three different units kJ/mol, kJ/kg, cal/g Your text has a table of Latent Heats, Melting and Boiling Points (18-2) Latent heats depend on p and T. Latent heat fusion heat vaporization (H 2 O) 79.7 cal/g 540 cal/g (1 atm) 333.5 kJ/kg 2257 kJ/kg (1 atm)

12 First Law of Thermodynamics Applied to the Human Body Food enters the body containing chemical energy Some is converted to stored chemical energy and some to thermal energy Then chemical energy is converted to work (mechanical energy) and some heat is released Sometimes the body has to do work to replace heat lost (shivering) If the internal energy of the body is constant then food energy must equal heat loss and work done What happens if not? Food Chemical Energy Work Heat Thermal Energy

13 Results of U Changes in the internal energy result in changes in the measurable macroscopic variables of the system Pressure Temperature Volume For the human body it is usually temperature or volume (isobaric)

14 Metabolic Rates (Cal/m 2 -hr) Sleeping35 Lying awake40 Sitting50 Standing60 Walking140 Running600 Shivering 250 Your surface area can be approximated using the formula SA =.202m.425 x h.725 where m is in kg and h is in meters. Calculate your surface area The metabolic rate at rest is the basal metabolic rate. The surface area of a 70 kg man of height 1.55m is about 1.70 m 2. His metabolic rate is therefore 40 x 1.70 = 68 Cal/hr while lying awake.

15 Heat and Life We need energy to function (blood circulation, cell repair, etc.) Even at rest a 70 kg person consumes about 70 Cal/hr The energy needed depends on a persons weight and build However, it has been found that human energy consumption (usage) divided by a person s surface area is approximately the same for most people It is given a unit of Cal/m 2 -hr and called metabolic rate

16 Measuring Metabolic Rate The metabolic rate is related to oxygen consumption by About 80 W is the basal metabolic rate, just to maintain and run different body organs

17 Various Metabolic Rates

18 Aerobic Fitness One way to measure a persons physical fitness is their maximum capacity to use or consume oxygen

19 Energy Output vs. Food Intake Food requirements depend on activity levels Consider this schedule: ActivityEnergy (Cal/m 2 ) For a person of SA 1.7m 2 this is 2320(1.70) 8 hr of sleep 280 or 3940 Cal per day. 8 hr of moderate activity 1200 This could be met by a 4 hr of reading 240diet of: 1 hr of heavy exercise 300 400g of carbs 1600 Cal 3 hr of dressing, eating 300 200g of Protein 800 Cal Total 2320171g of fat1540 Cal 3940 Cal

20 Weight Gain vs. Weight Loss They are stored as tissue (fat or muscle) Lack of caloric intake results in the body getting energy from stored fat first (9 Cal/g) and then proteins (4 Cal/g) The average person can go 50 days without food Angus Barbieri of Scotland consumed only tea, coffee and water from June 1965 to July 1966 reducing his body weight from 472 lbs to 178 lbs Pregnant women need an extra 136 Cal/day which can come from an increased appetite or a decrease in physical activity The body cannot eliminate excess calories

21 Efficiency of the Human Body Efficiency is the ratio of the mechanical power supplied to the metabolic rate or total power input

22 Example Hiking Howards Knob Suppose one starts out from King Street and climbs Howards Knob. How much energy is needed? Δh=1400 ft=427 m Assume m=80 kg PE= mgh = 80 kg 9.8 (m/s 2 ) 427 m = 335 kJ/4.186kj/kcal = 80 kcal ε=.2 ΔPE(ME)= εΔPE food ΔPE food =400 kcal

23 HEAT FLOW IN MATERIALS Heating occurs by Radiation, Convection and Conduction. In materials it is by conduction. ΔQ/Δt = κ A ΔT/L where is κ is the thermal conductivity. The unit of conductivity w/(mK),see text Table 18-3 for κ values.

24 HEAT FLOW IN MATERIALS Materials which have poor thermal conductivity can be thought of as thermal resistors. R = L/κ the metric unit of R is m 2 K /W in the English System R is ft 2 o F hr/BTU R(value)=1 ft 2 o F hr/BTU= 0.18 m 2 K/W R value of 4 inch fiberglass =11

25 HEAT FLOW IN MATERIALS R value for 6 inch fiberglass is =19 R value for 3/8 plasterboard is =0.32 R value for glass is = 0.02 R value for plywood = 0.6 R value for brick is between 0.6 1.0 If A = 1 ft 2, ΔT =(68-18) =50 o F, R=11 then, Q/t = 4.6 BTU/hr = 1.35 W.

26 HEAT FLOW IN MATERIALS In analogy with electrical phenomena one can apply the mathematical formalism of electrical phenomena to similar thermal phenomena. Electrical Thermal R R/A R in series R 1 /A + R 2 /A =R T /A

27 HEAT FLOW IN MATERIALS Electrical Thermal R in parallel A 1 /R 1 + A 2 /R 2 = A T /R eff Conductivity σ κ ρ (resistivity) = 1/σ 1/κ Δ V Δ T i = ΔQ Charge /Δt ΔQ Heat /Δt Flux i = Δ V/R ΔQ Heat /Δt = ΔT A/R

28 HEAT FLOW IN MATERIALS Let Electrical Conductivity σ = 1/ρ The ratio of the thermal conductivity and the electrical conductivity and the Kelvin Temperature is the Widermann- Franz –LorentzRatio σ/κT = 3 (R/F) 2 = 2.23 x 10 -8 (V/K) 2 Approximately constant for metals.

29 Work in a THERMAL SYSTEM Consider a piston of area A being pushed outward by a gas in a volume V The WORK DONE by the gas is: dW = F dx = p A dx = p dV W = V1 V2 p dV Sign of work determined by V 1 > or < V 2

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