1. Thermal Energy

2. Temperature & Heat Temperature is related to the average kinetic energy of the particles in a substance.

3. SI unit for temp. is the Kelvin K = °C + 273 (10°C = 283K) °C = K – 273 (10K = -263°C) Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance.

4. Thermal energy relationships As temperature increases, so does thermal energy (because the kinetic energy of the particles increased). Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy).

5. Heat The flow of thermal energy from one object to another. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler Cup gets cooler while hand gets warmer

6. Specific Heat Some things heat up or cool down faster than others. Land heats up and cools down faster than water

7. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). C water = 4184 J / kg C C sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

8. Specific Heat Which take longer to heat to 100°C? C (cal/g°C) C (J/g°C) Water1.004.18 Aluminum0.220.90 Copper0.0930.39 Silver0.0570.24 Gold0.0310.13 50g Al50g Cu Aluminum has a higher specific heat, so it will take longer to heat up. It will ALSO take longer to cool down.

9. Why does water have such a high specific heat? Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them. watermetal

10. How to calculate changes in thermal energy q = mCp  T q = change in thermal energy (heat) m = mass of substance  T = change in temperature (T f – T i ) C p = specific heat of substance -q means heat loss+q = heat gain

11. Heat Transfer A 32g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon? GIVEN m = 32g T i = 60°C T f = 20°C q = ?? C = 0.235 J/g°C WORK q = mCΔT m = 32g ΔT = T f - T i ΔT = 20°C – 60°C = -40°C q = (32g)(0.235J/g°C)(-40°C) q = -300.8 J

12. Heat Transfer How much heat is required to warm 230 g of water from 12°C to 90°C? GIVEN m = 230g T i = 12°C T f = 90°C Q = ?? C = 4.184 J/g°C WORK q = mCΔT m = 230g ΔT = T f - T i ΔT = 90°C – 12°C = 78°C q = (230g)(4.184J/g°C)(78°C) q = 75,061 J

13. A piece of iron at a temperature of 145°C cools off to 45°C. If the iron has a mass of 10 g and a specific heat of 0.449 J/g°C, how much heat is given up? GIVEN m = 10g T i = 145°C T f = 45°C q = ?? C = 0.449 J/g°C WORK q = mCΔT m = 10g ΔT = T f - T i ΔT = 45°C – 145°C = -100°C q = (10g)(0.449J/g°C)(-100°C) q = -449 J

14. A calorimeter is used to help measure the specific heat of a substance. Heat gained = Heat lost First, mass and temperature of water are measured Then heated sample is put inside and heat flows into water  T is measured for water to help get its heat gain This gives the heat lost by the substance Knowing its q value, its mass, and its  T, its C p can be calculated

15. Let’s Practice A 55.1 g piece of metal is heated to a temp of 45.1°C, and placed into a cup containing 359g of water at 20.0°C. The final temp of the water and metal is 22.3°C. How much heat energy did the water absorb? q = mcΔT q = (359g)(4.18J/g°C)(22.3°C – 20.0°C) = 3.45 x 10 3 J How much heat energy did the metal release to the water? q lost = q gained q lost by the metal = - 3.45 x 10 3 J The q is negative because heat was lost. What is the specific heat of the metal? 3.45 x 10 3 J = (55.1g)(C)(22.3°C – 45.1°C) 2.75 J/g°C = C