Heat Chapter 14 =2% of the AP Test according to CollegeBoard.com
Summary
 Heat is the energy that flows from the higher temperature object to the lower temperature object when the two are placed in contact.
 When two objects are in contact and are isolated from the environment, the heat gained by one is equal to the heat lost by the other; this is simply the conservation of energy.
 Unit: 1 calorie (cal)= the amount of heat necessary to raise the temperature of one gram of water by one Celsius degree. 4.186 Joules (the SI unit for heat)= 1 cal
 Temperature: measure of the average kinetic energy of all molecules in the object
 Internal Energy: total energy of all molecules in the object
 Internal energy of ideal monatomic gas (U)= (3/2)*nRT
 Heat: transfer of energy from obe object to another because of a differnce in temperature
Specific Heat
_

Calorimetry

Latent Heat
_

Derivations
heat_conduction_derivations.pptx  
File Size:  773 kb 
File Type:  pptx 
Demonstration 1
_SPECIFIC HEAT  ALUMINUM AND COPPER
PURPOSE: To illustrate calorimitry and to determine experimentally
the specific heats of aluminum and of copper.
DESCRIPTION: Boil the aluminum and copper blocks in a pan of water and place the metal blocks at 100 degrees C into equal amounts of water at room temperature in two styrofoam containers. Stir the water in the containers with the digital thermometer probe, measuring the temperature when equilibrium is reached. From these measurements the specific heats can be determined. Multiply the specific heat by the molar mass to find the molar specific heat. The values for both materials are nearly the same and equal to 3R=6(1/2)kT.
EQUIPMENT: Copper and aluminum 1 kg masses, styrofoam buckets each with 1 kG of roomtemperature water, pan balance, hot plate with pan and boiling water, hook for lifting hot masses, digital thermometer with probe.
SETUP TIME: 10 min.
DESCRIPTION: Boil the aluminum and copper blocks in a pan of water and place the metal blocks at 100 degrees C into equal amounts of water at room temperature in two styrofoam containers. Stir the water in the containers with the digital thermometer probe, measuring the temperature when equilibrium is reached. From these measurements the specific heats can be determined. Multiply the specific heat by the molar mass to find the molar specific heat. The values for both materials are nearly the same and equal to 3R=6(1/2)kT.
EQUIPMENT: Copper and aluminum 1 kg masses, styrofoam buckets each with 1 kG of roomtemperature water, pan balance, hot plate with pan and boiling water, hook for lifting hot masses, digital thermometer with probe.
SETUP TIME: 10 min.
Demonstration 2
_CONVECTION  HOT PLATE
PURPOSE: To see convection currents.
DESCRIPTION: The irregular refraction patterns created by convection currents in air heated from below are easily seen when a light from a projector shines through the air onto a screen. This phenomenon is often seen when the sun shines brightly onto surfaces like cars and roads, and is responsible for the twinkling of stars.
EQUIPMENT: Hot plate and point source of light.
SETUP TIME: 5 min..
DESCRIPTION: The irregular refraction patterns created by convection currents in air heated from below are easily seen when a light from a projector shines through the air onto a screen. This phenomenon is often seen when the sun shines brightly onto surfaces like cars and roads, and is responsible for the twinkling of stars.
EQUIPMENT: Hot plate and point source of light.
SETUP TIME: 5 min..
Demonstration 3
_TRANSFORMATION OF MECHANICAL ENERGY INTO HEAT
PURPOSE: To demonstrate transformation of mechanical energy into
heat.
DESCRIPTION: Use an electric drill to spin a wooden dowel rod in a hole on a large wooden beam. Shortly the contact point begins to smoke, indicating generation of heat due to friction.
EQUIPMENT: Quarterinch electric drill, 1/4" dowel rod, soft wood block.
SETUP TIME: 5 min.
DESCRIPTION: Use an electric drill to spin a wooden dowel rod in a hole on a large wooden beam. Shortly the contact point begins to smoke, indicating generation of heat due to friction.
EQUIPMENT: Quarterinch electric drill, 1/4" dowel rod, soft wood block.
SETUP TIME: 5 min.
Demonstration 4
_LATENT HEAT  ICE TO WATER TO STEAM
PURPOSE: To show latent heat as ice is transformed to water and
then to steam.
DESCRIPTION: A flask is filled to within one inch of the brim with a mixture of water and ice cubes at the freezing temperature of water. The flask is then heated for about 15 to 20 minutes with the burner on high, with the temperature measured by the dial thermometer. A plot of temperature as a function of time clearly shows that extra heat is required to produce the icewater and watersteam phase transitions.
EQUIPMENT: Hot plate, Erlenmeyer flask with cold water and ice cubes, clamp, and large dial thermometer.
SETUP TIME: 10 min.
DESCRIPTION: A flask is filled to within one inch of the brim with a mixture of water and ice cubes at the freezing temperature of water. The flask is then heated for about 15 to 20 minutes with the burner on high, with the temperature measured by the dial thermometer. A plot of temperature as a function of time clearly shows that extra heat is required to produce the icewater and watersteam phase transitions.
EQUIPMENT: Hot plate, Erlenmeyer flask with cold water and ice cubes, clamp, and large dial thermometer.
SETUP TIME: 10 min.
Demonstration 5
_PULSE GLASS
PURPOSE: To illustrate change of phase between liquid and vapor
in an confined fluid.
DESCRIPTION: Hold the pulse glass so that the bulbs point up, and enclose one bulb with your hand to warm it. The heat from your hand will change some of the liquid on that end to vapor, increasing the pressure in that end and forcing the liquid into the other end.
EQUIPMENT: Pulse glass.
SETUP TIME: None.
DESCRIPTION: Hold the pulse glass so that the bulbs point up, and enclose one bulb with your hand to warm it. The heat from your hand will change some of the liquid on that end to vapor, increasing the pressure in that end and forcing the liquid into the other end.
EQUIPMENT: Pulse glass.
SETUP TIME: None.
Demonstration 6
_THERMAL EXPANSION  BIMETAL STRIP
PURPOSE: To demonstrate differential thermal expansion.
DESCRIPTION: Strips of invar steel and brass are welded together to form a bimetal strip. Because their coefficients of thermal expansion are different, heating the bimetal strip over a burner results in bending, as illustrated. This type of material is used in some types of thermostatic controllers.
EQUIPMENT: Bimetal strip with burner.
SETUP TIME: None.
DESCRIPTION: Strips of invar steel and brass are welded together to form a bimetal strip. Because their coefficients of thermal expansion are different, heating the bimetal strip over a burner results in bending, as illustrated. This type of material is used in some types of thermostatic controllers.
EQUIPMENT: Bimetal strip with burner.
SETUP TIME: None.
Demonstration 7
_THERMAL EXPANSION  BALL AND RING
PURPOSE: To demonstrate thermal expansion.
DESCRIPTION: When both ball and ring are at room temperature, the ball fits through the ring. If only the ball is heated, it expands so that it will not fit through the ring.
EQUIPMENT: Ball and ring on handles with burner.
SETUP TIME: None.
DESCRIPTION: When both ball and ring are at room temperature, the ball fits through the ring. If only the ball is heated, it expands so that it will not fit through the ring.
EQUIPMENT: Ball and ring on handles with burner.
SETUP TIME: None.
AP Multiple Choice Questions with Solutions
_(1) When water is heated from 0º C to 20º C its volume
(a) goes on increasing
(b) goes on decreasing
(c) remains constant up to 15º C and then increases
(d) first decreases and then increases
(e) remains constant up to 4º C and then increases
This question is meant just for checking your knowledge of the behaviour of water. Water has maximum density at nearly 4º C and hence the correct option is (d).
(2) 5 g of ice at 0º C is mixed with 10 g of water at 10º C. The temperature of the mixture is
(a) 0º C
(b) 2º C
(c) 2.5º C
(d) 5º C
(e) 7.5º C
In order to melt 5 g of ice (into water) without change of temperature 400 calories of heat are required since the latent heat of fusion for icewater change is nearly 80 calories per gram. The heat that is released by 10 g of warm water at 10º C on cooling to 0º C is 100 calories only since the specific heat of water is 1 calorie per gram per Kelvin.
So the warm water can melt just a quarter of the amount of ice and the mixture will remain at 0º C [Option (a)].
(3) Equal masses of three liquids of specific heats C1, C2and C3 at temperatures t1, t2and t3 respectivelyare mixed. If there is no change of state, the temperature of the mixture is
(a) (t1+ t2+ t3)/3
(b) (C1t1+ C2t2+ C3t3)/[3(C1+ C2+ C3)]
(c) (C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(d) 3(C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(e) 3(t1+ t2+ t3)
If the mass of each liquid is m, the total amount of heat (H) initially is given by
H = m(C1t1+ C2t2+ C3t3)
After mixing the same amount of heat is available. If the common temperature is t, we have
H = mt(C1+ C2+ C3)
From the above equations, t = (C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(4) The amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is (universal gas constant = R)
(a) 2 R
(b) 3 R
(c) 5 R
(d) 5R/2
(e) 7R/2
The molar specific heat of a mono atomic ideal gas at constant pressure (cp) is 5R/2 where R is the universal gas constant.
[The molar specific heat of a mono atomic ideal gas at constant volume (cp) is 3R/2 and in accordance with Meyer’s relation, cp = cv + R].
Therefore, the amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is 1×(5R/2) ×2 = 5R.
(5) Two identical rectangular strips, one of copper and the other of steel, are riveted as shown to form a bimetal strip. On heating, the bimetal strip will
(a) get twisted
(b) remain straight
(c) bend with steel on the convex side
(d) bend with steel on the concave side
(e) contract
On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side.
[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].
(6) Four cylindrical rods of different radii and lengths are used to connect two heat reservoirs at fixed temperatures t1 and t2 respectively. From the following pick out the rod which will conduct the maximum quantity of heat:
(a) Radius 1 cm, length 1 m
(b) Radius 1 cm, length 2 m
(c) Radius 2 cm, length 4 m
(d) Radius 3 cm, length 8 m
(e) Radius 0.5 cm, length 0.5 m
The quantity of heat conducted is directly proportional to the area of cross section and inversely proportional to the length of the rod (when the same temperature difference exists between the ends).
[Remember that the quantity of heat Q = KAdθ/dx where K is the thermal conductivity, A is the area of cross section and dθ/dx is the temperature gradient].
When you compare rod (a) with rod (b) you find that rod (a) can conduct better since its length is less than that of rod (b).
Rod (a) and rod (c) conduct equally since the cross section area as well as the length of rod (c) is 4 times that of rod (a).
Rod (d) is better than rod (a) since its cross section area is 9 times that of rod (a) while its length is only 8 times that of rod (a).
Rod (e) is worse than rod (a) since its cross section area is a quarter of that of rod (a) while its length is half that of rod (a).
Therefore, the rod which will conduct the maximum quantity of heat is (d).
[It is enough to compare the ratio of the area to length. The area is directly proportional to the square of the radius. Since the unit of radius is centimetre and that of length is metre in all cases, you can blindly compare the values of 12/1, 12/2, 22/4, 32/8 and (0.5)2/(0.5). The highest value is 32/8].
(a) goes on increasing
(b) goes on decreasing
(c) remains constant up to 15º C and then increases
(d) first decreases and then increases
(e) remains constant up to 4º C and then increases
This question is meant just for checking your knowledge of the behaviour of water. Water has maximum density at nearly 4º C and hence the correct option is (d).
(2) 5 g of ice at 0º C is mixed with 10 g of water at 10º C. The temperature of the mixture is
(a) 0º C
(b) 2º C
(c) 2.5º C
(d) 5º C
(e) 7.5º C
In order to melt 5 g of ice (into water) without change of temperature 400 calories of heat are required since the latent heat of fusion for icewater change is nearly 80 calories per gram. The heat that is released by 10 g of warm water at 10º C on cooling to 0º C is 100 calories only since the specific heat of water is 1 calorie per gram per Kelvin.
So the warm water can melt just a quarter of the amount of ice and the mixture will remain at 0º C [Option (a)].
(3) Equal masses of three liquids of specific heats C1, C2and C3 at temperatures t1, t2and t3 respectivelyare mixed. If there is no change of state, the temperature of the mixture is
(a) (t1+ t2+ t3)/3
(b) (C1t1+ C2t2+ C3t3)/[3(C1+ C2+ C3)]
(c) (C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(d) 3(C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(e) 3(t1+ t2+ t3)
If the mass of each liquid is m, the total amount of heat (H) initially is given by
H = m(C1t1+ C2t2+ C3t3)
After mixing the same amount of heat is available. If the common temperature is t, we have
H = mt(C1+ C2+ C3)
From the above equations, t = (C1t1+ C2t2+ C3t3)/ (C1+ C2+ C3)
(4) The amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is (universal gas constant = R)
(a) 2 R
(b) 3 R
(c) 5 R
(d) 5R/2
(e) 7R/2
The molar specific heat of a mono atomic ideal gas at constant pressure (cp) is 5R/2 where R is the universal gas constant.
[The molar specific heat of a mono atomic ideal gas at constant volume (cp) is 3R/2 and in accordance with Meyer’s relation, cp = cv + R].
Therefore, the amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is 1×(5R/2) ×2 = 5R.
(5) Two identical rectangular strips, one of copper and the other of steel, are riveted as shown to form a bimetal strip. On heating, the bimetal strip will
(a) get twisted
(b) remain straight
(c) bend with steel on the convex side
(d) bend with steel on the concave side
(e) contract
On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side.
[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].
(6) Four cylindrical rods of different radii and lengths are used to connect two heat reservoirs at fixed temperatures t1 and t2 respectively. From the following pick out the rod which will conduct the maximum quantity of heat:
(a) Radius 1 cm, length 1 m
(b) Radius 1 cm, length 2 m
(c) Radius 2 cm, length 4 m
(d) Radius 3 cm, length 8 m
(e) Radius 0.5 cm, length 0.5 m
The quantity of heat conducted is directly proportional to the area of cross section and inversely proportional to the length of the rod (when the same temperature difference exists between the ends).
[Remember that the quantity of heat Q = KAdθ/dx where K is the thermal conductivity, A is the area of cross section and dθ/dx is the temperature gradient].
When you compare rod (a) with rod (b) you find that rod (a) can conduct better since its length is less than that of rod (b).
Rod (a) and rod (c) conduct equally since the cross section area as well as the length of rod (c) is 4 times that of rod (a).
Rod (d) is better than rod (a) since its cross section area is 9 times that of rod (a) while its length is only 8 times that of rod (a).
Rod (e) is worse than rod (a) since its cross section area is a quarter of that of rod (a) while its length is half that of rod (a).
Therefore, the rod which will conduct the maximum quantity of heat is (d).
[It is enough to compare the ratio of the area to length. The area is directly proportional to the square of the radius. Since the unit of radius is centimetre and that of length is metre in all cases, you can blindly compare the values of 12/1, 12/2, 22/4, 32/8 and (0.5)2/(0.5). The highest value is 32/8].
Free Response/Multiple Choice intermixed
test_questions  
File Size:  13 kb 
File Type:  docx 
test_answers  
File Size:  34 kb 
File Type:  docx 
AP FreeResponse with Solutions
#6

#3

_Essentials for multiple choice questions
_(1) Temperature tC in Celsius scale can be converted to temperature tF in Fahrenheit scale using the relation tC/100 = (tF – 32)/180 Remember that the ice point and steam point of water are 32º F and 212º F respectively in the Fahrenheit scale. These temperatures in the Celsius scale are 0º C and 100º C respectively so that a temperature difference (∆tF) of 180º in the Fahrenheit scale is equal to a temperature difference (∆tC) of 100º in the Celsius scale. The significance of the numbers 100, 32 and 180 will be clear to you now.
(2) Temperature tC in Celsius scale can be converted to temperature tK in Kelvin scale (absolute scale) using the relation tC/100 = (tK – 273.15)/100
so that tC = tK – 273.15 or, tK = tC + 273.15. Often the above relation is written as tK = tC + 273 very nearly. The above equation follows from the fact that the ice point and steam point of water are 273.15 K and 373.15 K respectively in the Kelvin scale. A temperature difference of 1º in the Kelvin scale is equal to a temperature difference of 1º in the Celsius scale.
(3) The increase in length (∆ℓ) of a solid on raising its temperature by ∆T is given by ∆ℓ = α ℓo ∆Twhere α is the coefficient oflinear expansion (linear expansivity)and ℓo is the original length.
(4) The increase in area (∆A) of a solid on raising its temperature by ∆T is given by ∆A = β Ao ∆Twhere β is the coefficient ofarea expansion (area expansivity)and Ao is the original area. In the case of isotropic homogeneous solids β = 2α
(5) The increase in volume (∆V) of a solid on raising its temperature by ∆T is given by ∆V = γVo ∆Twhere γ is the coefficient ofvolume expansion (volume expansivity)and Vo is the original volume. In the case of isotropic homogeneous solids γ = 3α
(6) Specific heat capacity (specific heat) C of a substance is the quantity of heat absorbed or rejected by 1 kg of the substance to change the temperature by 1 K. Therefore, the heat involved in changing the temperature of m kg of a substance through ∆TK (or ∆T º C) is mC ∆Twhere C is the specific heat of the substance.
(7) Molar specific heat of a substance is the quantity of heat absorbed or rejected by 1 mole of the substance to change the temperature by 1 K.You know that in the case of gases there are two specific heats viz., specific heat at constant volume and specific heat at constant pressure. Similarly there are two molar specific heats: molar specific heat at constant volume and molar specific heat at constant pressure.
(8) Heat transfer takes place by three processes: conduction, convection and radiation. The quantity of heat Q conducted in a time t through a uniform rod of length L and area of cross section A when the ends of the rod are maintained at a temperature difference ∆Tis given by Q = KA (∆T/L) t where K is the thermal conductivity of the material of the rod [The temperature gradient∆T/L is often written as ∆T/∆x, replacing L by ∆x, which will be more appropriate in the case of heat conduction through slabs of thickness ∆x]. The rate of transfer of heat by conduction (time rate), H is given by H = Q/t = KA∆T/L According to Stefan’s law the energy (E) radiated per second per unit surface area of a perfectly black body is directly proportional to the 4th power of the absolute temperature of the body: E α T4 Or, E = σT4 where σ is Stefan’s constant.
So if the temperature is doubled, the energy radiated from the body will become 16 times. Newton’s law of cooling says that the rate of cooling (rate of loss of heat) of a body is directly proportional to the excess of temperature of the body over the surroundings: dQ/dt α (T2 – T1) where dQ is the heat lost in a time dt when the temperatures of the body and the surroundings are respectively T2 and T1. Note that the loss of heat mentioned in Newton’s law of cooling is due to all the three mechanisms viz., conduction, convection and radiation. Further, the excess temperature (T2 – T1) should be small.
(2) Temperature tC in Celsius scale can be converted to temperature tK in Kelvin scale (absolute scale) using the relation tC/100 = (tK – 273.15)/100
so that tC = tK – 273.15 or, tK = tC + 273.15. Often the above relation is written as tK = tC + 273 very nearly. The above equation follows from the fact that the ice point and steam point of water are 273.15 K and 373.15 K respectively in the Kelvin scale. A temperature difference of 1º in the Kelvin scale is equal to a temperature difference of 1º in the Celsius scale.
(3) The increase in length (∆ℓ) of a solid on raising its temperature by ∆T is given by ∆ℓ = α ℓo ∆Twhere α is the coefficient oflinear expansion (linear expansivity)and ℓo is the original length.
(4) The increase in area (∆A) of a solid on raising its temperature by ∆T is given by ∆A = β Ao ∆Twhere β is the coefficient ofarea expansion (area expansivity)and Ao is the original area. In the case of isotropic homogeneous solids β = 2α
(5) The increase in volume (∆V) of a solid on raising its temperature by ∆T is given by ∆V = γVo ∆Twhere γ is the coefficient ofvolume expansion (volume expansivity)and Vo is the original volume. In the case of isotropic homogeneous solids γ = 3α
(6) Specific heat capacity (specific heat) C of a substance is the quantity of heat absorbed or rejected by 1 kg of the substance to change the temperature by 1 K. Therefore, the heat involved in changing the temperature of m kg of a substance through ∆TK (or ∆T º C) is mC ∆Twhere C is the specific heat of the substance.
(7) Molar specific heat of a substance is the quantity of heat absorbed or rejected by 1 mole of the substance to change the temperature by 1 K.You know that in the case of gases there are two specific heats viz., specific heat at constant volume and specific heat at constant pressure. Similarly there are two molar specific heats: molar specific heat at constant volume and molar specific heat at constant pressure.
(8) Heat transfer takes place by three processes: conduction, convection and radiation. The quantity of heat Q conducted in a time t through a uniform rod of length L and area of cross section A when the ends of the rod are maintained at a temperature difference ∆Tis given by Q = KA (∆T/L) t where K is the thermal conductivity of the material of the rod [The temperature gradient∆T/L is often written as ∆T/∆x, replacing L by ∆x, which will be more appropriate in the case of heat conduction through slabs of thickness ∆x]. The rate of transfer of heat by conduction (time rate), H is given by H = Q/t = KA∆T/L According to Stefan’s law the energy (E) radiated per second per unit surface area of a perfectly black body is directly proportional to the 4th power of the absolute temperature of the body: E α T4 Or, E = σT4 where σ is Stefan’s constant.
So if the temperature is doubled, the energy radiated from the body will become 16 times. Newton’s law of cooling says that the rate of cooling (rate of loss of heat) of a body is directly proportional to the excess of temperature of the body over the surroundings: dQ/dt α (T2 – T1) where dQ is the heat lost in a time dt when the temperatures of the body and the surroundings are respectively T2 and T1. Note that the loss of heat mentioned in Newton’s law of cooling is due to all the three mechanisms viz., conduction, convection and radiation. Further, the excess temperature (T2 – T1) should be small.