Unit 10 TEST REVIEW

1. Unit 10 TEST REVIEW

2. Vocabulary • Temperature • Average kinetic energy of the atoms or molecules of an object • Hotter objects have higher kinetic energy. • Thermal Energy • Total energy of an object. • Heat • Thermal energy of a material that moves from warmer to cooler object • Measured in Joules

3. Types of Heat Transfer • Conduction • Heat transfer between colliding particles when objects touch • Convection • Heat transfer in a fluid • Radiation • Heat transfer through a vacuum by electromagnetic waves

4. Heat Gained or Lost Q = m C ΔT m = Q / (C ΔT) C = Q / (m ΔT) ΔT = Q / (m C) ΔT = (TF –TI) Q – heat [Joules or Calories] (+) Q heat heat gained (-) Q heat lost m – mass [grams or kilograms] C – specific heat [J / kg °C] [J / kg K] [cal / g °C] ΔT – change in temperature [°C or K] Look at ALL units for problems

5. Specific Heat • The amount of heat required to raise the temperature of mass of a material by 1 unit of temperature • Units • J / kg °C • J / kg K • cal / g °C

6. Temperature Conversions

7. +Qm = -Qw • When a hot metal is placed in a colder water, heat energy is transferred from hot to cold. None of the energy is lost. You can solve for the initial temperature of the metal or the final temperature of them both.

8. Specific Heat / Temperature Conversions Example Problem Solutions

9. Problem 1 Q = mCΔT Mass 0.34 kg Specific Heat 14300 J/kg K Change in Temperature 25 K Q = (0.34 kg) (14300 J/kg K) (25 K) Q= 121550 J 1.22 E5 J

10. Problem 4 C = Q / (m ΔT) Mass 0.59 kg TI 98°C Heat Energy 21,100 J TF 6.8°C C = 21,100 J / (0.59 kg  91.2°C) C = 392 J /kg°C

11. Problem 7 m = Q /(CΔT) Heat Energy 9900000 J Change in Temperature 55K Specific Heat 1000 J/kg K m = 9900000 J / (1000 J/kg K  55K) m = 180 kg

12. Problem 9 ΔT = Q /(mC) ΔT = TF – TI Mass 0.225 kg Specific Heat 2300 J/kg K Heat Energy 3900 J TF 18°C (291 K) 291 K – TI = -3900 J / (0.225kg  2300 J /kg K) 291 K – TI = -7.54 K TI = 298.5 K