Multiple Choice Practice Questions on Heat Transfer and Thermal Expansion for AP Physics B

“Genius is one percent inspiration and ninety nine percent perspiration”
– Thomas A. Edison

Today we will discuss some practice questions (MCQ) on heat transfer and thermal expansion. You may click here to obtain the essential points you need to note in this section.
Here are the questions:
(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 ice-water 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, C₂ and C₃ at temperatures t₁, t₂ and t₃ respectively are 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 bi-metal strip. On heating, the bi-metal 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].

AP Physics B - Multiple Choice Practice Questions on Kinetic Theory of Gases

Essential points to be remembered in kinetic theory of gases were discussed in the post dated 13^th March 2008. Questions on kinetic theory of gases were discussed subsequently. You can access all posts related to kinetic theory of gases by clicking on the label, ‘kinetic theory’ below this post. To access older posts you need to click on the ‘older posts’ button.

Today we will discuss a few more typical multiple choice questions on kinetic theory of gases:

(1) The root mean square (R.M.S.) speed v of the molecules of an ideal gas is given by the expressions,

v = √(3RT/M ) and

v = √(3kT/m ) where R is universal gas constant, T is the absolute (Kelvin) temperature, M is the molar mass, k is Boltzman’s constant and m is the molecular mass. The R.M.S. speed of oxygen molecules (O₂) at temperature T₁ is v₁. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).

(a) v₁/2

(b) v₁

(c) √2 v₁

(d) 2v₁

(e) 4v₁

We have v₁ = √(3RT₁/M ) or

v₁ = √(3kT₁/m )

On dissociation the molar mass as well as the molecular mass gets halved. Using the second equation, the R.M.S. speed v after dissociation is given by

v = √[3k×2T₁/ (m/2 )] = 2√(3kT₁/m ) = 2v₁

(2) Four moles of an ideal diatomic gas is heated at constant volume from 20º C to 30º C. The molar specific heat of the gas at constant pressure (C_p) is 30.3 Jmol^–1K^–1 and the universal gas constant (R) is 8.3 Jmol^–1K^–1. The increase in internal energy of the gas is

(a) 80.3 J

(b) 303 J

(c) 332 J

(d) 880 J

(e) 1212 J

The increase in internal energy is MC_v∆T where M is the mass of the sample of the gas, C_v is the specific heat at constant volume and ∆T is the rise in temperature of the gas. If we use the molar specific heat of the gas at constant volume for C_v, the number of moles in the sample of the gas is to be used in the place of M.

Now, C_v = C_p – R = 30.3 – 8.3 = 22 Jmol^–1K^–1.

Therefore, the increase in internal energy of the gas is 4×22×10 = 880 J.

(3) In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to

(a) intermolecular attraction and the size of molecules

(b) size of molecules and expansion of the container

(c) expansion of the container and intermolecular attraction

(d) kinetic energy of molecules and collision of molecules

(e) intermolecular attraction and collision of molecules

In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for P arises due to intermolecular attraction.

The entire volume V of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.

The correct option is (a).

(4) Gases exert pressure on the walls of the container because the gas molecules

(a) collide one another

(b) exert intermolecular attraction

(c) possess momentum

(d) expand on absorbing heat

(e) exert repulsive force

Because of the momentum of the gas molecules, they collide with the walls of the containing vessel and momentum transfer takes place, resulting in a force on the walls. Pressure is force per unit area. The basic reason for the pressure is the momentum of the gas molecules [Option (c)].

Now, see similar questions with solution here.

Temperature and Heat for AP Physics B- Equations to be Remembered

Here are the essential things you need to remember for solving multiple choice questions involving heat and temperature:

(1) Temperature t_C in Celsius scale can be converted to temperature t_F in Fahrenheit scale using the relation

t_C/100 = (t_F – 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 (∆t_F) of 180º in the Fahrenheit scale is equal to a temperature difference (∆t_C) of 100º in the Celsius scale. The significance of the numbers 100, 32 and 180 will be clear to you now.

(2) Temperature t_C in Celsius scale can be converted to temperature t_K in Kelvin scale (absolute scale) using the relation

t_C/100 = (t_K – 273.15) /100

so that t_C = t_K – 273.15 or, t_K = t_C + 273.15.

Often the above relation is written as t_K = t_C + 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 ℓ_o ∆T where α is the coefficient of linear expansion (linear expansivity) and ℓ_o is the original length.

(4) The increase in area (∆A) of a solid on raising its temperature by ∆T is given by

∆A = β A_o ∆T where β is the coefficient of area expansion (area expansivity) and A_o 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 = γV_o ∆T where γ is the coefficient of volume expansion (volume expansivity) and V_o 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 ∆T K (or ∆T º C) is mC ∆T where 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 viz., molar specific heat at constant volume and molar specific heat at constant pressure. In this context you will find a useful post here.

(8) Heat transfer takes place by three processes viz., 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 ∆T is 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 4^th 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.

In the next post we will discuss questions in this section. Meanwhile find a useful post here.

AP Physics B – Answers to Practice Questions (MCQ) on Kinetic Theory

Some typical multiple choice questions for practice were given to you in the post dated 15^th March 2008. As promised, I give below the answers with explanation.

(1) If a diatomic gas molecule has an additional vibrational mode which contributes to both kinetic and potential energies, the sum of which is equal to kT where k is Boltzmann’s constant and T is the absolute temperature, what is the ratio of specific heats of the gas?

(a) 3/2

(b) 4/3

(c) 5/3

(d) 7/5

(e) 9/7

A diatomic molecule without vibrational mode has average energy equal to (5/2)kT since it has 5 degrees of freedom. Since the vibrational mode contributes an extra energy of kT, the total average energy per molecule is (5/2)kT+kT = (7/2)kT.

The molar specific heat at constant volume (C_v) is therefore equal to (7/2)kT×N where N is Avogadro’s number.

Therefore, C_v = (7/2)R and the molar specific heat at constant pressure (C_p) is [(7/2)R+R] = (9/2)R where R is universal gas constant.

Therefore, the ratio of specific heats, γ = C_p/C_v = 9/7

(2) Four moles of helium (a mono atomic gas) is contained in a barrel provided with a light, frictionless piston (fig.). The quantity of heat to be supplied to the gas for raising its temperature by 5 K is (universal gas constant, R = 8.31 J mol^–1 K^–1)

(a) 415.5 J

(b) 249.3 J

(c) 136.5 J

(d) 166.2 J

(e) 581.7 J

When heat is supplied to the gas, the internal energy of the gas is increased, raising the temperature of the gas. The pressure of the gas is unchanged since the piston moves outwards, doing work against the atmospheric pressure. It is the molar specific heat at constant pressure (C_p) that is involved here so that the quantity of heat (Q) to be supplied is given by

Q = no. of moles × molar specific heat at constant pressure × rise in temperature.

Helium being mono atomic, molar specific heat at constant pressure (C_p) is (5/2)R

Therefore, Q = 4×(5/2)R×5 = 4×(5/2)×8.31×5 = 415.5 J

(3) The intermolecular attraction is an important reason why real gases behave differently compared to ideal gas. But real gases too can behave like ideal gas at

(a) low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and high pressure

(d) high temperature and low pressure

(e) absolute zero (0 K) temperature

At high temperatures and low pressures the molecules are far apart and molecular interactions are negligible. Under these conditions even a real gas behaves like an ideal gas [Option (d)].

(4) A fixed mass of an ideal gas at pressure P is contained in a closed vessel of volume V. It is heated so that the root mean square velocity of the gas molecules is doubled. The thermal expansion of the vessel is negligible. Then the increase in pressure (∆P) of the gas is

(a) P

(b) 2P

(c) 3P

(d) 4P

(e) √2 P

The root mean square velocity of the gas molecules is given by c = √(3kT/m) where k is Boltzmann’s constant, T is the absolute temperature and m is the mass of the molecule. The r.m.s. velocity is therefore directly proportional to the square root of absolute temperature. Since the r.m.s. velocity is doubled on heating the gas, the temperature must be quadrupled.

By Charle’s law, we have PV/T = constant for a given mass of gas. When the temperature is quadrupled at constant volume, the pressure must be quadrupled. The final pressure is thus 4P and the increase in pressure is 3P.

(5) A cubical vessel of side ℓ contains N molecules of an ideal gas. If the mass of a molecule is m and the root mean square velocity is c, the pressure of the gas is

(a) mNc² /3ℓ³

(b) mNc² /ℓ³

(c) mc² /3ℓ³

(d) 3mNc² /ℓ³

(e) (1/3)mNc²

The pressure exerted by a gas is given by P = (1/3) ρc² where ρ is the density of the gas and c is the root mean square velocity of the gas molecules. The density of the gas here is

ρ = Total mass/Total volume = mN/ℓ³

Therefore, pressure P = mNc² /3ℓ³

(6) Four moles of oxygen (diatomic) is mixed with eight moles of neon (mono atomic). If they are treated as ideal gases, the effective molar specific heat at constant volume of the mixture is (R = universal gas constant)

(a) (5/3) R

(b) (7/5) R

(c) (9/5) R

(d) (11/6) R

(e) (13/8) R

The molar specific heat at constant volume (C_v) of a diatomic gas is (5/2)R and that of a mono atomic gas is (3/2)R where R is universal gas constant.

Molar specific heat is the heat required to raise the temperature of one mole of the gas by 1 K. The quantity of heat required to raise the temperature of four moles of oxygen and eight moles of neon by 1 K is [4×(5/2)R + 8×(3/2)R] = 22 R and the total number of moles in the mixture is 12.

Therefore, molar specific heat of the mixture = 22 R/12 = (11/6) R

(7) The temperature at which the root mean square velocity of hydrogen gas molecules is twice that at 0º C is

(a) 20º C

(b) 40º C

(c) 819º C

(d) 859º C

(e) 1132º C

Since the root mean square velocity is directly proportional to the square root of absolute temperature, it will be doubled at 4 times the absolute temperature corresponding to 10º C. Therefore, the answer is 4×283 K = 1132 K = (1132 – 273) K = 859º C.

(8) Equal number of oxygen molecules (molar mass m₁) and helium molecules (molar mass m₂) are kept in two identical vessels. If they are at the same temperature, their pressures will be in the ratio

(a) m₁/m₂

(b) m₂/m₁

(c) √(m₁/m₂)

(d) √(m₂/m₁)

(e) 1

Most of you know that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. If equal numbers of molecules of different gases have the same temperature and volume their pressure must therefore be the same. The correct option is (e).

The expression for pressure in the form, P = nkT also is handy in working out similar problems. Since k is Boltzmann’s constant, the pressure depends only on the temperature and the number density (number per unit volume). You can easily obtain this expression by substituting for the r.m.s. velocity [c = √(3kT/m)] in the expression for pressure, P = (1/3) nmc².

AP Physics B – Practice Questions (MCQ) on Kinetic Theory

As promised in the post dated 13^th March 2007, I give below some typical multiple choice questions on kinetic theory for practice. For the time being I leave these questions here without solution.

(1) If a diatomic gas molecule has an additional vibrational mode which contributes to both kinetic and potential energies, the sum of which is equal to kT where k is Boltzmann’s constant and T is the absolute temperature, what is the ratio of specific heats of the gas?

(a) 3/2

(b) 4/3

(c) 5/3

(d) 7/5

(e) 9/7

(2) Four moles of helium (a mono atomic gas) is contained in a barrel provided with a light, frictionless piston (fig.). The quantity of heat to be supplied to the gas for raising its temperature by 5 K is (Universal gas constant, R = 8.31 J mol^–1 K^–1)

(a) 415.5 J

(b) 249.3 J

(c) 136.5 J

(d) 166.2 J

(e) 581.7 J

(3) The intermolecular attraction is an important reason why real gases behave differently compared to ideal gas. But real gases too can behave like ideal gas at

(a) low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and high pressure

(d) high temperature and low pressure

(e) absolute zero (0 K) temperature

(4) A fixed mass of an ideal gas at pressure P is contained in a closed vessel of volume V. It is heated so that the root mean square velocity of the gas molecules is doubled. The thermal expansion of the vessel is negligible. Then the increase in pressure (∆P) of the gas is

(a) P

(b) 2P

(c) 3P

(d) 4P

(e) √2 P

(5) A cubical vessel of side ℓ contains N molecules of an ideal gas. If the mass of a molecule is m and the root mean square velocity is c, the pressure of the gas is

(a) mNc² /3 ℓ³

(b) mNc² /ℓ³

(c) mc² /3ℓ³

(d) 3mNc² /ℓ³

(e) (1/3)mNc²

(6) Four moles of oxygen (diatomic) is mixed with eight moles of neon (mono atomic). If they are treated as ideal gases, the effective molar specific heat at constant volume of the mixture is (R = universal gas constant)

(a) (5/3) R

(b) (7/5) R

(c) (9/5) R

(d) (11/6) R

(e) (13/8) R

(7) The temperature at which the root mean square velocity of hydrogen gas molecules is twice that at 10º C is

(a) 20º C

(b) 40º C

(c) 819º C

(d) 859º C

(e) 1132º C

(8) Equal number of oxygen molecules (molar mass m₁) and helium molecules (molar mass m₂) are kept in two identical vessels. If they are at the same temperature, their pressures will be in the ratio

(a) m₁/m₂

(b) m₂/m₁

(c) √(m₁/m₂)

(d) √(m₂/m₁)

(e) 1

Try to answer these questions. The essential points required for solving these questions were discussed in the post dated 13^th March 2008, which you can access by clicking on the label ‘kinetic theory’ below this post.

I’ll be back with the solution in a day or two.