Heating of air • Energy (e.g. solar radiation) absorbed by an air parcel will lead to temperature change Stull (3.1) Q  mairc pT –cp = 1005 J/kg/K for dry air –cp = 4200 J/kg/K for liquid water

– Note: for moist air, cp depends on the mixing ratio r (Stull equation 3.2). However, the value differs from that for dry air by less than a few percent. In this class we will ignore this dependence Numerical Examples • 5000 J of energy is supplied to heat 2 kg of dry air. What is the temperature change?

– T = Q/mair/cp = 5000/2/1005 = 2.5 K • If the same amount of energy is supplied to heat 2 kg of water, what is the temperature change?

– T = Q/mwater/cp = 5000/2/4200 = 0.6 K • If the same amount of energy is supplied to

heat 2 kg of granite (cp = 794 J/kg/K), what is the temperature change?

– T = Q/m/cp = 5000/2/794 = 3.1 K Latent heat • When a mass m of water vapor condenses or water freezes, latent heat is released Q  m  L Stull (3.3)

• Similarly, to evaporate a mass m of water or to melt a mass m of ice, latent heat has to be supplied Latent heat values • Latent heat of – Vaporization (liquid -> vapor): -2.5 x 106 J/kg – Melting (solid -> liquid): -3.3 x 105 J/kg – Sublimation (solid -> vapor): -2.83 x 106 J/kg

– Condensation (vapor -> liquid): 2.5 x 106 J/kg – Freezing (liquid -> solid): 3.3 x 105 J/kg – Deposition (vapor -> solid): 2.83 x 106 J/kg

– Negative means heat is absorbed; positive means heat is released Numerical Examples

• 5000 J of energy is supplied to an ice cube to melt the ice. How much ice is melted? – m = Q/L = 5000/3.3x105 = 0.015 kg = 15 g Classwork Example • A moist air parcel contains 1 kg of dry air and 3 g of water vapor. If all the vapor condenses, how much warmer will the air parcel become? – Condensation of water vapor releases latent heat, the energy released acts to warm the air parcel – Latent heat released Q = L m = 2.5x106x0.003 = 7500 J – When 7500 J of energy is supplied to heat 1 kg of dry air

• T = Q/m/cp = 7500/1/1005 = 7.5 K First Law of Thermodynamics • As we saw earlier, heating can change the temperature of an air parcel • Temperature can also be changed if work is done on or by the air parcel – e.g. air released from a tire expands and does work and cools • The equation governing temperature change, including the effects of work done: Q P st  c p T  -- 1 Law of Thermodynamics m  air Stull (3.4) Adiabatic lapse rate • As an air parcel rises (subsides), it expands (is compressed), and work is done by (done on) it, hence its temperature changes as it rises (subsides) • Using the Hydrostatic equation p = -gz the 1st law can be rewritten as g Q T   z  Stull (3.5) c p mair c p • If there is no heat exchange between the air parcel and the environment, Q = 0 (adiabatic) • Hence, as an air parcel rises adiabatically, it does work on the environment and cools, the adiabatic lapse rate is T  g     9.8K / km  10 K / km Stull (3.6) z  adiabatic c p