Heat Heat Capacity „ Heat capacity connects heat flow to temperature change: „ What is heat? Q = CΔT ‰ Heat (Q) is the “flow” or “transfer” of energy from ‰ Heat capacity C depends on material, and also on the quantity of one system to another material present. Eliminate quantity dependence by introducing ‰ Often referred to as “heat flow” or “heat transfer” specific heat c and molar heat capacity c′:

‰ Requires that one system must be at a higher Q = mcΔT Q = nc′ΔT m = mass n = number of moles temperature than the other „ Heat will only flow from the system with the higher Latent Heat temperature to the system with the lower temperature „ A phase change occurs when a solid melts to a liquid, a liquid boils to „ Heat will only flow from the system with the higher average a gas, a gas condenses to a liquid, and a liquid freezes to a solid. internal energy to the system with the lower average ‰ Each of these phase changes requires a certain amount of heat, internal energy although the temperature does not change. ‰ If a solid becomes liquid, or vice versa, the amount of heat per gram is „ Total internal energy does not matter. the latent heat of fusion. ‰ If a liquid becomes gas, or vice versa, the amount of heat per gram is the latent heat of vaporization

A glass is filled with 100 g of ice at 0.00°C and 200 g of water at 25.0°C. Latent Heat (a) Characterize the content of the glass after equilibrium has been reached. Neglect heat transfer to and from the environment. (b) Repeat your calculations for 50.0 g of ice and 250 g of water.

1 Work Work Done by Thermal Systems

Work can be done by thermal systems, as in the expansion of a gas.

Using the definitions of work and pressure:

V W = 2 PdV ∫V1

Note, work can be done on a thermal system, as in the compression of a gas.

First Law of Thermodynamics Types of Transformations

„ When temperature changes, internal energy has changed – „ Isobaric, ΔP = 0 may happen through heat transfer or through mechanical work ‰ W = PΔV „ First law is a statement of conservation of energy „ Work = Pressure*Change in Vol „ Change in internal energy of system equals the difference between the heat added to the system and the work done by the system

„ ‰ Differential form

‰ Heat added +, heat lost -, work done by system +, work done on system – „ Isochoric, ΔV = 0

‰ Internal Energy U is a state property ‰ W = 0 ⇒ΔU = Q ‰ Work W and heat Q are not „ ‰ But work and heat are involved in thermodynamic processes that change the state of the system

2 The PV diagram shows two states of a system containing

1.45 moles of an ideal gas (P1 = P2 = 450 Types of Transformations 2 3 3 N/m , V1 = 2.00 m , V2 = 8.00 m ). A) Draw an isobaric process from state 1 to state 2. „ Isothermal, ΔT = 0 B) Draw a two-step process that depicts and isothermal expansion from state 1 to V2 followed by an isochoric ‰ ΔU = 0, ⇒ W = Q increase in temperature to state 2. C) In both cases, calculate the work done, the heat added „ or lost, and the change in internal energy.

„ Adiabatic, Q = 0

‰ ⇒ΔU = -W

„

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Molar Specific Heats for Gasses Equipartition of Energy „ Molar specific heats for gasses are different if heat is added at constant pressure vs constant volume

‰ QP = nCPΔT ‰ QV = nCVΔT If the two processes result „ Isobaric, ΔP = 0 in the same temperature change, ΔU is the same. ‰ „ Diatomic, triatomic, etc. molecules are more complex ‰ ‰ Molecules can translate, rotate, and vibrate „ Isochoric, ΔV = 0 ‰ Energy is shared equally between the various degrees of freedom ‰

3 o A certain gas has a specific heat cV = 0.0356 kcal/kg- C, which changes little over a wide temperature range. What is the atomic mass of the gas? What gas is it? Adiabatic, Q = 0 Assume an adiabatic and quasistatic expansion of an ideal gas.

After a lot of calculus and algebra (see p 592):

For the same increase in volume, an adiabatic process will result in a lower pressure and lower temperature than an isothermal process.

What about work?

Heat Transfer Conduction

„ Conduction „ Time rate of heat transfer depends on ‰ Results from molecular interactions ‰ The material „ Specifically k = thermal conductivity „ Collisions? ‰ Area ‰ Energy is transferred through interaction ‰ Temperature difference „ Convection ‰ Thickness or length ‰ Results from the mass transfer of material ‰ Think fluid flow „ Radiation Differential form R-Value, Thermal Resistance Value ‰ Energy transferred by electromagnetic radiation (waves) ‰ Does not require a “medium”

4 Your refrigerator can be thought of as a box with six sides of total area 2.5 m2. The effective R value of the walls is 1.5 m2–K/W. The temperature inside is 5.0°C, Radiation while the temperature outside is 25°C. Calculate the rate of heat loss. „ Time rate of heat transfer depends on

‰ The material „ Specifically e = emissivity

‰ Area „ Surface area

‰ Temperature difference „ Experimentally determined to be proportional to the 4th power of T

‰ An experimentally determined Stefan-Blotzmann constant σ

A surface that is “white hot” emits about 10 times more power than a “red hot” surface. What does this tell us quantitatively about the relative temperature? Blackbody Radiation „ Objects that become sufficiently hot will glow visibly; as they get hotter they go from red, to yellow, to a bluish white. „ This is electromagnetic radiation; objects at any temperature will emit it at various frequencies, from radio waves all the way to gamma rays. „ This radiation from a body in thermal equilibrium is called blackbody radiation, as it is purely thermal and doesn’t depend on any properties of the body other than its temperature and area. „ Deriving the energy density as a function of frequency and temperature required introducing some new concepts:

‰ Here, c is the speed of light:

‰ And h is Planck’s constant:

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