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. 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.
Temperature Scales Around the world, three temperature scales are used: Fahrenheit (oF), Celsius (oC), Kelvin (K)
Temperature Scales 212 What is the freezing point and boiling point of water on the Fahrenheit scale? 32
Temperature Scales 100 What is the freezing point and boiling point of water on the Celsius scale?
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!)
Temperature Scales 373 What is the freezing point and boiling point of water on the Kelvin scale? 273
Temperature Conversions How to convert from Celsius to Fahrenheit: Fahrenheit temp. = (1.8 x Celsius temp.) +32.0 TF = 1.8 TC + 32.0
Temperature Conversions How to convert from Fahrenheit to Celsius: Celsius temp. = (Fahrenheit temp. - 32.0) 1.8 TC = (TF - 32.0)
Temperature Conversions How to convert from Celsius to Kelvin and vice versa: Celsius to Kelvin: TK = TC + 273 Kelvin to Celsius: TC = TK - 273
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
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
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
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
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
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
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.
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.
Heat Transfer The warmer object always transfers heat to the cooler object. The greater the temperature difference, the faster the energy transfer.
Heat Transfer Three methods of energy transfer are conduction, convection, and radiation.
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.
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.
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).
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.
Conductors and Insulators Insulators are a material that is a poor energy conductor. Wood, rubber, fiberglass, Styrofoam, wool, etc.
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.
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.
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)
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)
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 * m * ΔT) c = 1.3 J/g*oC 10 g
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
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