 # The second law of thermodynamics: The heat flow statement: Heat flows spontaneously from a substance at a higher temperature to a substance at a lower.

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The second law of thermodynamics: The heat flow statement: Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction.

A heat engine is a device that uses heat to perform work. The efficiency e of a heat engine is the ratio of work W done to the input heat Q H. Efficiency is often stated as a percentage.

For a heat engine the total of the work W done and the rejected heat Q C must equal the input heat Q H. Q H = W + Q C

Ex 6 - An automobile has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?

A reversible process is one in which both the system (the gas) and its environment (the piston and the rest of the universe) can be returned to exactly the states they were in before the process occurred. A process that involves an energy-dissipating mechanism cannot be reversible.

Sadi Carnot proposed that a heat engine has maximum efficiency when the processes within the engine are reversible.

Carnot’s Principle: No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency.

Notice that this law does not state that a reversible engine is 100% efficient. Carnot’s law is an alternative statement of the second law of thermodynamics.

A Carnot engine is an idealized reversible engine. The Kelvin scale is based on the comparative temperatures of the reservoirs of a Carnot engine.

The ratio of rejected heat Q C to input heat Q H is: Q C / Q H = T C / T H. T C and T H must be expressed in Kelvins.

The efficiency of a Carnot engine is equal to 1 - T C /T H.

Ex 7 - A heat engine has a hot reservoir at 298.2 K and a cold reservoir at 280.2 K. (a) Find the maximum possible efficiency for such an engine. (b) Determine the minimum input heat Q H that would be needed if a number of these heat engines were to produce an amount of work equal to the 9.3 x 10 19 J of energy that the US consumed in 1994.

The efficiency is low because the temperatures are so close to being the same. The efficiency increases as T C approaches absolute zero. Experiments have shown that it is not possible to cool a substance to 0 K; therefore a 100% efficient heat engine is not possible.

Refrigerators, air conditioners, and heat pumps use work to force heat to flow from the cold reservoir to the hot reservoir; this is called a refrigeration process. Q H = W + Q C

If the refrigeration system is an ideal device, then Q C /Q H = T C /T H applies, just as for a Carnot engine.

Q H = W + Q C and Q C /Q H = T C /T H are the two equations used for Carnot refrigeration systems.

Ex 8 - An ideal or Carnot heat pump is used to heat a house to a temperature of T H = 294 K. How much work must be done by the pump to deliver Q H = 3350 J of heat into the house when the outdoor temperature TC is (a) 273 K and (b) 252 K?

Irreversible processes cause a machine to lose some of its ability to perform work. This partial loss is expressed in terms of entropy. Heat divided by temperature is the change in entropy for a reversible system: ∆S = (Q/T) R. The subscript R means “reversible” system.

Reversible processes do not alter the total entropy of the universe. ∆S universe = 0 for a reversible process.

Ex 11 - 1200 J of heat flows spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur.

Any irreversible process increases the entropy of the universe. ∆S universe > 0 for an irreversible process. The entropy of the universe continually increases.

The second law of thermodynamics stated in terms of entropy - The total entropy of the universe does not change when a reversible process occurs (∆S universe = 0) and increases when an irreversible process occurs (∆S universe > 0).

Ex 12 - Suppose 1200 J of heat is used as input for a n engine under two different conditions. In the first, the heat is supplied by a hot reservoir whose temperature is 650 K. In the other, the heat flows irreversibly through a copper rod into a second reservoir whose temperature is 350 K and then enters the engine. In either case, a 150-K reservoir is used as the cold reservoir. For each case, determine the maximum amount of work that can be obtained from the 1200 J of heat.

The previous problem showed that 240 J less work could be performed by the system with a lower temperature difference. Example 11 showed that the entropy of the universe increased by 1.6 J/K.

W unavailable = T 0 ∆S universe In the previous cases, W unavailable = 240 J, T 0 = the cold reservoir temp of 150 K, ∆Suniverse = 1.6 J/K; so: W unavailable = 150 K 1.6 J/K = 240 J.

An increase in entropy is associated with an increase in disorder.

The third law of thermodynamics - It is not possible to lower the temperature of any system to absolute zero in a finite number of steps.

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