Presentation on theme: "AP Physics Chp 15. Thermodynamics – study of the relationship of heat and work System vs Surroundings Diathermal walls – allow heat to flow through Adiabatic."— Presentation transcript:
Thermodynamics – study of the relationship of heat and work System vs Surroundings Diathermal walls – allow heat to flow through Adiabatic walls – do not allow heat to flow
Zeroth Law of Thermodynamics Two systems in thermal equilibrium with a third are also in equilibrium with each other
First Law of Thermodynamics Internal energy changes based on the amount of heat and/or work done by/on the system. U = Q – W W = PV Q is positive when it goes in (endothermic) W is positive when the system does work
What is the change in the internal energy if you supply 15 kJ to a 35 m 3 sample of helium at 101150 Pa and it is allowed to expand to 52 m 3 ?
U = Q – W U = 15000 J – (101150 Pa)(52 m 3 – 35 m 3 ) U =
If a process is slow enough then the P and T are uniform. When P is constant its called an isobaric process. W = PV Why is W negative when work is done on a system?
Isochoric processes occur at constant volume This is the bomb calorimeter idea.
At constant T its an isothermal process Adiabatic processes occur without the transfer of any heat
One way to relate work for a system is to plot the P vs V graph and compare the area under the curve.
How much work is done in compressing the gas from 4 m 3 to 3 m 3 ? Why is it more than 9 m 3 to 8 m 3 ?
What would a graph for an isochoric process look like? Why does it show no work being done?
Efficiency can be multiplied by 100 to make it a percentage. Since Q H = W + Q C W = Q H – Q C e = 1 – Q C /Q H
Carnot created a principle that says that a irreversible engine can not have a greater efficiency than a reversible one operating at the same temperatures. For a Carnot engine Q C /Q H = T C /T H e carnot = 1 – T C /T H
If absolute zero could be maintained while depositing heat in then a 100% efficiency would be possible but its not.
If my truck operates at a running temperature of 94 o C and the outside air is only -5 o C, what is the maximum efficiency for the engine?
T H = 273 +94 = 367 K T C = 273 + -5 = 268 K e = 1 – T C /T H e = 1 – 268K / 367 K = 0.27 or 27%
Refrigerators, Air Conditioners, Heat Pumps All of these take heat from the cold reservoir and put it into the hot reservoir by doing a certain amount of work. Its the reverse of the heat engine.
Why cant you cool your house by running an air conditioner without having it exhaust outside? Coefficient of performance = Q C /W Heat pumps warm up a space by moving heat from the cold outside to the warm inside.
Seems kind of weird that the cold outside has heat. If you use a Carnot heat pump to deliver 2500 J of heat to your house to achieve a temperature of 20 o C while it is -5 o C outside, how much work is required?
W = Q H – Q C and Q C /Q H = T C /T H So Q C = Q H T C /T H and W = Q H – Q H T C /T H W = Q H (1-T C /T H ) W = 2500J (1- 268 K/293K) = 210 J
Entropy Randomness or disorder gas>>>liquids>solids The entropy of the universe increases for irreversible process but stays constant for reversible