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Heat and Temperature

Heat and Temperature. Chapter 14. Temperature. Temperature is a measure of the average kinetic energy of all the particles within an object. We measure temperature using a thermometer. Temperature. Thermometers work on a principle called thermal expansion .

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Heat and Temperature

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  1. Heat and Temperature Chapter 14

  2. Temperature • Temperature is a measure of the average kinetic energy of all the particles within an object. • We measure temperature using a thermometer.

  3. Temperature • Thermometers work on a principle called thermal expansion. • As the temperature rises, the particles of a substance gain kinetic energy and move faster. With this increased motion, the particles of that substance move further apart and expand the volume of a substance.

  4. Temperature Scales • Around the world, three temperature scales are used: Fahrenheit (oF), Celsius (oC), Kelvin (K)

  5. Temperature Scales 212 • What is the freezing point and boiling point of water on the Fahrenheit scale? 32

  6. Temperature Scales 100 • What is the freezing point and boiling point of water on the Celsius scale? 0

  7. Temperature Scales • The Kelvin scale is based on the principle of absolute zero. • Absolute zero is the coldest possible temperature, and it is also where an object’s energy is minimal. • Absolute zero is -273.13 oC. The Kelvin scale was created so that absolute zero was in fact recorded as 0 kelvin or 0 K. (Kelvin has no negatives!)

  8. Temperature Scales 373 • What is the freezing point and boiling point of water on the Kelvin scale? 273

  9. Temperature Conversions • How to convert from Celsius to Fahrenheit: Fahrenheit temp. = (1.8 x Celsius temp.) +32.0 TF = 1.8 TC + 32.0

  10. Temperature Conversions • How to convert from Fahrenheit to Celsius: Celsius temp. = (Fahrenheit temp. - 32.0) 1.8 TC = (TF - 32.0) 1.8

  11. Temperature Conversions • How to convert from Celsius to Kelvin and vice versa: Celsius to Kelvin: TK = TC + 273 Kelvin to Celsius: TC = TK - 273

  12. Temperature Conversions • If the boiling point of liquid hydrogen is -252.87 oC, what is its boiling point in oF? TF = 1.8 TC + 32.0 TF = 1.8 (-252.87) + 32.0 TF = -423.17 oF

  13. Temperature Conversions • If the temperature of a winter day at the North Pole is -40.0 oF, what is the temperature in oC? TC = (TF - 32.0) / 1.8 TC = (-40.0- 32.0) / 1.8 TC = -40.0 oC

  14. Temperature Conversions • The melting point of gold is 1064 oC, what is its melting point in K? TK = TC + 273 TK = 1064 + 273 TK = 1337 K

  15. Temperature Conversions • The air temperature in a typical living room is 294 K, what is its temperature in oC? TC = TK – 273 TC = 294 K – 273 TC = 21 oC

  16. Temperature Conversions • The metal in a running car engine will typically get as hot as 388 K, how hot is this in oF? TC = TK – 273 TC = 388 K – 273 TC = 115 oC TF = TC1.8+ 32.0 TF = (115 oC) 1.8+ 32.0 TF = 239 oF

  17. Temperature Conversions • The air temperature on a summer day in the desert is typically 110 oF, what is the temperature in K? TC = (TF - 32.0) / 1.8 TC = (110 oF - 32.0) / 1.8 TC = 43 oC TK = TC + 273 TK = 43 oC + 273 TK = 316 K

  18. Heat • “Hot and “cold” has nothing to do with temperature. • “Cold” is felt when we lose energy to an object with less kinetic energy. • “Hot” is felt when we gain energy from an object with more kinetic energy.

  19. Heat • Heat is the transfer of thermal energy from the particles of one object to those of another object due to a temperature difference between two objects.

  20. Heat Transfer • The warmer object always transfers heat to the cooler object. • The greater the temperature difference, the faster the energy transfer.

  21. Heat Transfer • Three methods of energy transfer are conduction, convection, and radiation.

  22. Heat Transfer • Thermal conduction is the transfer of energy as heat between particles as they collide within a substance or between two objects in contact.

  23. Heat Transfer • Convection is the transfer of energy by the movement of fluids with different temperatures. • A convection current is the flow of a fluid due to heated expansion followed by cooling and contraction.

  24. Heat Transfer • Radiation is the transfer of energy by electromagnetic waves. • Radiation does not require contact. • It is the only method of energy transfer that can occur in a vacuum (outer space).

  25. Conductors and Insulators • Conductors are a material through which energy can be easily transferred as heat. • Solids, especially metals like copper and silver, are the best conductors.

  26. Conductors and Insulators • Insulators are a material that is a poor energy conductor. • Wood, rubber, fiberglass, Styrofoam, wool, etc.

  27. Laws of Thermodynamics • First Law of Thermodynamics: the total energy used in any process is conserved, whether the energy is transferred as a result of work, heat, or both.

  28. Laws of Thermodynamics • Second Law of Thermodynamics: the energy transferred as heat always moves from an object at a higher temperatures to an object at a lower temperature.

  29. Specific Heat Capacity • Specific heat capacity is the amount of energy transferred as heat that will raise the temperature of 1 kg of a substance by 1 K. • Specific heat capacity is a physical property and will be the same any pure substance. Therefore, we can use it to identify substances. energy = specific heat capacity x mass x change of temp. q = c x m x ΔT (J) (kg) (K) (J/kg * K)

  30. Specific Heat Capacity • Specific heat capacity is the amount of energy transferred as heat that will raise the temperature of 1 kg of a substance by 1 K. (J) (kg) (K) (J/kg * K)

  31. Ch. 14 Section 2 Energy Transfer • A 10 g piece of iron absorbs 1000 joules of heat energy, and its temperature changes from 25°C to 100°C. Calculate the specific heat capacity of iron. q = ΔT= c = m = 1000 J c = q / (m * ΔT) c = 1000J/(10g*75oC) 100-25 = 75 oC q c = 1000J/(750g*oC) ? c = 1.3 J/g*oC (c * m * ΔT) 10 g

  32. Ch. 14 Section 2 Energy Transfer • To what temperature will a 20 g piece of glass raise if it absorbs 1000 joules of heat and its specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 10.0°C. ΔT = q / (c * m) ΔT = 1000J/(0.50 J/goC *20g) q = ΔT= c = m = 1000 J ΔT = 1000/(10 oC) Tf-10 = ? q ΔT = 100 oC 0.50 J/goC ΔT = Tf – To (c * m * ΔT) 100oC = Tf – 10 oC 20 g 10oC+100oC=Tf –10oC+10oC Tf = 110oC

  33. Ch. 14 Section 2 Energy Transfer • 100 g of 5.0°C water is heated until its temperature is 20°C. If the specific heat of water is 4.18 J/g°C, calculate the amount of heat energy needed to cause this rise in temperature. q = ΔT= c = m = ? q = c * m * ΔT q = 4.18 J/goC * 100 g * 15oC 15 oC q q = 6,270 J 4.18 J/goC (c * m * ΔT) 100 g

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