1. Thermal Flow If you can’t stand the heat

2. Temperature  As we know Temperature is the average kinetic energy of the molecules. As they bounce around they exchange energy.  Transfer of heat between liquids and gas’s are easy to see since the molecules are relatively free to move. But for a solid the energy transfer is less about it flying around and more about it’s vibration.

3. Balancing it all out.  The flow of energy in any system is always from high to low. Just like when we pour water on a hill the water wants to go to the lowest point it can.  The same can be said of heat. Heat will always travel from ‘hot’ to ‘cold’ and never the other way around.  It will flow untill it is ‘balanced’ or ‘at equilibrium’ with it’s surroundings. (e.g. the water flowing down the hill ended up in a lake. The water can not go lower down the hill)

4. Flow of heat  Convection: When heat flows between gas or liquid  As the air or the liquid increases in temperature the ‘hotter’ portion moves upwards (hot air rises) away from the heating source and cooling where it falls back down.  Example: your convection oven.

5. Heat flow  Conduction: the transfer of heat from direct contact between 2 objects.

6. Heat flow  Radiation: does not use matter to transfer the heat. Think of the Sun heating the Earth. There is nothing between the Earth and the Sun but we still feel its warmth.

7. Thermal energy conservation  Just like with momentum and energy thermal energy can be conserved.  So if we take 2 blocks one at ‘hot’ and one at ‘cold’ what do you think will happen?

8. Heat flow  We will get ‘warm’  The change in the first blocks heat will equal the change in the 2 nd blocks heat just in reverse.  So basically Q lost = Q gain  Or as the law of conservation states, there should be no loss so they should sum up to zero.  Q 1 + Q 2 = 0 (remember one is losing heat so it will be negative)

9. Example  A 29.5kg sample of methanol at 208.9 K is mixed with 54.3kg of methanol at 302.3 K. Calculate the final tempeerature if the specific heat of methanol is 2.53 J/kgK  (mass) (Δt) (C p ) = (mass) (Δt) (C p )  Q lost on the left; Q gain on the right.  Substituting and solving, we have:  (29.5) (x - 208.9) (2.53) = (54.3) (302.3 - x) (2.53)  29.5x - 6162.55 = 16414.89 - 54.3x  83.8x = 22577.44  X = 269.4K

10. Background  This method was used to determine the specific heat of different object.  They would take a calorimeter and put in the substance. Then measure the change in the temperature.