Calorimetry (Heat Transfer) | iCalculator™

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Thermodynamics Learning Material
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13.4Calorimetry (Heat Transfer)

In this Physics tutorial, you will learn:

  • How does the process of heat transfer occur?
  • What does the general law of calorimetry say?
  • Which are the four methods of heat transfer?
  • What is a calorimeter used for?
  • How can we calculate the rate of heat transfer in each of the four methods of heat transfer?


What does a blacksmith do when he wants to cool down a piece of hot processed iron? Does he wait until the iron cool down by itself or he uses any fast method for this? Explain.

What do you obtain if mix hot water and cold water? Is it relevant the amount of each water sample?

Suggest a method to measure the heat supplied to a substance.

How many methods of heat transfer do you know? What do they have in common? Different?

This tutorial will focus on the measurement of heat exchange between thermodynamic systems. In scientific terminology, this process is known as "calorimetry". Also, the methods of heat transferred and their calculation will be extensively discussed. Let's have a closer look at all these.

Heat Transfer

As explained in the previous tutorial "Absorption of Heat. States of Matter. Change of State.", when two substances are placed in contact, there is a heat flow between them. The direction of heat flow is pre-determined; heat always flows from the hottest to the coldest object.

This "heat flow" does not necessarily imply any matter transfer; rather, it often takes place when the boundary atoms of the two objects collide with each other at different speeds. Such a collision only makes the slow vibrating molecules of the cold object vibrate faster due to their collision with the fast vibrating molecules of the hot object, as discussed in the previous tutorials.

However, in many cases there is also some matter transfer during the process of heat exchange between systems. This occurs for example when we mix some hot and cold water. The result will be an amount of warm water, whose mass is the sum of the individual masses. Obviously, it is expected that the warm water have an in-between temperature.

The general law of calorimetry states that:

During a heat exchange process between two objects of a thermodynamic system, the heat released by the hottest object is entirely gained by the coldest object if the system is isolated from the surroundings.

Mathematically, we can write

Qreleased by the hot object = Qgained by the cold object

We can write for the heat released by the hottest object

Q1 = m1 × c1 × ∆t1
= m1 × c1 × (t1-tf )

where m1 is the mass of the hottest object, c1 is its specific heat capacity, t1 is the initial temperature of the hottest object and tf is the final (common) temperature after the heat exchange process is completed.

On the other hand, we can write for heat gained by the coldest object

Q2 = m2 × c2 × ∆t2
= m2 × c2 × (tf - t2 )

where m2 is the mass of the coldest object, c2 is its specific heat capacity, t2 is the initial temperature of the coldest object and tf is the final (common) temperature after the heat exchange process is done.

Pay attention to the expressions within the parenthesis. In the expression for the hottest object, we subtract the final temperature after heat exchange from the initial temperature as it gives a positive value, while in the expression for the coldest object we subtract the initial temperature before the contact from the final temperature in order to obtain a positive value. Therefore, combining the two above equations, we obtain the four equivalent equations (which basically represent the same thing):

Qreleased by the hot object = Qgained by the cold object


Q1 = Q2


m1 × c1 × ∆t1 = m2 × c2 × ∆t2


m1 × c1 × (t1 - tf ) = m2 × c2 × (tf - t2 )

Example 1

20 g water at 80°C is mixed with 30 alcohol at 10°C. Find the final temperature if no heat exchange with the surroundings does occur. Take the specific heat capacity of water as 1 cal/g°C and that of alcohol as 0.58 cal/g°C.

Solution 1

Using the notations introduced earlier in the theory, we write the following clues:

m1 = 20 g
m2 = 30 g
t1 = 80°C
t2 = 10°C
c1 = 1 cal/g°C
c2 = 0.58 cal/g°C
tf = ?

Applying the equation of heat exchange

m1 × c1 × (t1 - tf ) = m2 × c2 × (tf - t2 )

we obtain after substituting the known values:

20 × 1 × (80 - tf ) = 30 × 0.58 × (tf - 10)
20 × (80 - tf ) = 17.4 × (tf - 10)
1600 - 20 × tf = 17.4 × tf - 174
37.4 × tf = 1774
tf = 1774/37.4

We can plot a Temperature vs Heat graph like the one shown in the figure below to illustrate the phenomenon of heat exchange between two objects: one hot and one cold.

Physics Tutorials: This image provides visual information for the physics tutorial Calorimetry (Heat Transfer)

Example 3

Calculate the specific heat capacities of the two substances that exchange heat according the data provided in the graph below. Take the mass of the first substance as 2 kg and that of the second substance as 1 kg.

Physics Tutorials: This image provides visual information for the physics tutorial Calorimetry (Heat Transfer)

Solution 3

From the graph, we can extract the following clues:

t1 = 60°C
t2 = 20°C
tf = 45°C
Q1 = Q2 = 30 000 J

Also we have from the problem's words:

m1 = 2 kg
m2 = 1 kg
c1 = ?
c2 = ?

The heat released by the hot object is

Q1 = m1 × c1 × (t1 - tf )

Thus, the specific heat capacity of the hot object is

c1 = Q1/m1 × (t1 - tf )
= 30 000/2 × (60 - 45)
= 1000 J/kg°C

The heat absorbed by the cold object is numerically the same as the heat released by the hot object. Thus, we can write:

Q2 = m2 × c2 × (tf-t2 )

Therefore, the specific heat capacity of the cold object is

c2 = Q2/m2 × (tf - t2 )
= 30 000/1 × (45-20)
=1200 J/kg°C

What is a Calorimeter?

A calorimeter is an insulated container used to measure the specific heat capacity of an unknown solid by means of heat exchange approach. The only difference with the previous examples lies in the fact that the materials do not mix to obtain a homogenous mixture but they remain two distinct materials even after the thermal equilibrium is established (when both materials reach the same temperature). The rest is the same.

Example 4

A 20 g metal piece at 120°C is inserted into a calorimeter, which contains 200 g of water at 20°C. When thermal equilibrium is established, the final temperature of metal becomes 22°C. What is the specific heat capacity of metal if water has a specific heat capacity of 1 cal/g°C?

Solution 4

From the equation of heat exchange

Qreleased by the hot object = Qgained by the cold object

we can determine the value of specific heat capacity of metal, given that

m1 = 20 g
t1 = 120°C
m2 = 200 g
t2 = 20°C
tmix = 22°C
c1 = 1 cal/g°C
c2 = ?

Thus, substituting the above values in the equation

m1 × c1 × (t1 - tf ) = m2 × c2 × (tf - t2 )

we obtain

20 × c1 × (120 - 22) = 200 × 1 × (22 - 20)
20 × c1 × 98 = 200 × 2
1960 × c1 = 400
c1 = 400/1960
=0.204 cal/g°C

We can use the heat exchange method described above to calculate the final temperature even when one of the substances experiences any change in phase during the process. Let's consider an example.

Example 5

A 20 g ice cube at -5°C is inserted into a container, in which there are 100 g of water at 60°C. What will be the final temperature of the system after the thermal equilibrium is established? Take cice = 0.5 cal/g°C, cwater = 1 cal/g°C and Lice = 80 cal/g.

Solution 5

Most likely (given the difference in mass), the ice cube will experience three stages: 1- increase in temperature up to the melting point (for water it is 0°C), 2- melting process (without any change in temperature) and 3- increase in temperature of water up to the final temperature.

The general equation in the specific case is

mice × cice × ∆tice+mice × Lice+mformer ice(now cold water) × cwater × ∆tcold water=mhot water × cice × ∆tice

Substituting the values, we obtain

20 × 0.5 × [0 - (-5)]+20 × 80+20 × 1 × (tf - 0) = 100 × 1 × (60 - tf)
50 + 1600 + 20 × tf = 6000 - 100 × tf
20 × tf + 100 × tf = 6000 - 1600
120 × tf = 4400
tf = 36.7°C

Methods of Heat Transfer

Heat is transferred from one place to another through four basic methods. They are:

a) Conduction

As stated earlier, particles in hot objects vibrate faster than in cold objects. When we place in contact a hot and a cold object, heat flows through the collision of particles from the hot to the cold object. This method of heat transfer is known as conduction.

By definition, conduction is the method of heat transfer through the collision of particles.

In simpler words, conduction is the heat flow between objects through direct contact.

Physics Tutorials: This image provides visual information for the physics tutorial Calorimetry (Heat Transfer)

Not all object conduct the heat equally. Some materials conduct heat better than the others. As explained earlier, such materials as known as good conductors of heat. For example, most metals are good conductors of heat.

On the other hand, some materials conduct very poorly the heat. They are known as bad (poor) conductors of heat. Wood, glass, cork, plastics, air, vacuum etc., are all bad conductors of heat.

If you touch a wooden and a metal object inside a room in a cold winter day, the metal feels colder. This is not because metal has a lower temperature than wood. These two materials have the same temperature, as both they are within the same room. The reason why metal feels colder is because it is a better heat conductor than wood. Thus, when we touch the materials, heat flows at a higher rate from our body to metal than from our body to the wooden object. This process of heat flow takes place only in one direction - from our body to the material and not vice-versa because our body has a normal temperature of 37°C, which is much higher than that of a room temperature in winter.

Conduction is a method of heat transfer that does not involve any matter transfer between objects. However, matter must be present during this process, as it is needed to transmit the heat through collision between particles.

There is a mathematical equation, which allows us to calculate the rate of heat flow through a solid. Thus, the heat rate (heat per unit time) flowing through a solid object of surface A and thickness d during which the temperature changes from T1 to T2 is

Q/t = K × A × (T2-T1)/d

where K is a quantity known as the coefficient of thermal conduction, measured in [W/m × K]. It depends on the characteristics of material.

Example 6

What is the amount of heat absorbed by a 40 cm2 copper plate of 4 mm thickness if during 2 minutes the temperature of copper increases from 20°C to 30°C? Take the coefficient of thermal conduction for copper equal to 385 W/m × K.

Solution 6

In this problem, there are the following clues:

A = 40 cm2 = 0.004 m2
d = 4 mm = 0.004 m
t = 2 min = 120 s
T1 = 20°C
T2 = 30°C
K = 385 W/m × K
Q = ?

We have:

Q/t = K × A × (T2 - T1)/d
Q = K × A × (T2 - T1 )/d × t
= 385 × 0.004 × (30 - 20) × 120/0.004
= 462 000 J

b) Convection

When we mix two liquids or gases, the heat is distributed equally throughout the entire volume of mixture. This method of heat transfer is known as convection.

By definition, convection is the method of heat transfer through the circulation of fluid itself.

This process involves both heat and matter transfer. Obviously, matter must be present during convection.

There are two types of convection: natural and forced convection. Wind is an example of natural convection as it occurs without the intervention of humans. On the other hand, air flowing when we turn on an electric fan is an example of forced convection as this process involves human made tools to produce air circulation.

Coastal breeze is another example of natural convection. During the day, the sand heats up faster than water (water is one of substances with the highest specific heat capacity; this means water heats up and cools down much slower than other substances). Therefore, the air above the sand expands more than that above water. Such expansion causes a decrease in air density above the sand (air becomes rarer than before). The fresh air coming from regions above the sea replaces the missing air on the land regions. In this way, a slight wind called sea breeze is produced.

During the night, the reverse process does occur. Water is warmer than sand because it cools down slower. As a result, the air above water surface expands and it is replaced by fresh air coming from above the sandy regions. Therefore, a slight wind called land breeze is produced.

Physics Tutorials: This image provides visual information for the physics tutorial Calorimetry (Heat Transfer)

We can find the rate of heat flow through convection using the equation

Q/t = k × A × (T2 - T1 )

where k is the coefficient of thermal convection measured in [W/m2K] and A is the area involved.

Example 7

A fluid flows over a plane surface 2 m by 2 m. The surface temperature is 40°C, the fluid temperature is 30°C and the convective heat transfer coefficient is 1000 W/m2K. What is convective heat transfer between the surface and the fluid in 10 seconds?

Solution 7


A = 2 m × 2 m = 4 m2
T1 = 30°C
T2 = 40°C
k = 1000 W/m2
t = 10 s
Q = ?

Applying the equation of convective heat transfer

Q/t = k × A × (T2 - T1 )

we obtain after rearranging and substituting the known values:

Q = k × A × t × (T2 - T1 )
= 1000 × 4 × 10 × (40 - 30)
= 400 000 J

c) Radiation

Radiation is the third method of heat transfer. By definition, radiation is the method of heat transfer by means of electromagnetic waves.

Most radiation comes to Earth from the Sun. Radiation does not involve any matter transfer; it only transfers energy. Also, matter presence is not necessary in this process, as most radiation travels through vacuum.

When radiation falls on an object, it is partly reflected, partly transmitted and partly absorbed. The absorbed heat causes the molecules of the object to vibrate more than before, so it becomes hot.

Some objects emit radiation more than the others; they are known as black bodies. This is because black bodies absorb radiation more than other bodies. As a result, they accumulate a lot of energy. As a result, when black bodies become hot, they emit large amounts of this radiation to the surroundings.The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed with Stefan-Boltzmann Law as

Q/t=σ × T4 × A

where σ is a quantity known as Stefan-Boltzmann constant. It has a value of 5.6703 × 10-8 W/m2K4. T is the absolute temperature of the black body in Kelvin degree and A is the area of the emitting body.

Here, a black body is more an idealization than reality; it is quite impossible to find a perfect black body that is able to absorb all radiation falling on it and then, emitting all this amount of heat to the surroundings. For objects other than ideal black bodies ("gray bodies") the Stefan-Boltzmann Law can be expressed as

Q/t = ε × σ × T4 × A

where ε is a dimensionless quantity between 0 and 1 (one for black bodies) called emissivity coefficient of the object.

Example 8

A grey body of emissivity equal to 0.4 has an emitting surface of 3 m2. How many joules of heat energy per second does this body emit when it is at 27°C?

Solution 8

We have:

ε = 0.4
A = 3 m2
T = 27°C = 300 K = 3 × 102 K
σ = 5.6703 × 10-8 W/m2K4
Q/t = ?

Applying the Stefan-Boltzmann Law for grey bodies

Q/t = ε × σ × T4 × A

we obtain after substitutions

Q/t = 0.4 × 5.6703 × 10-8 × (3 × 102 )4 × 3
= 551 J/s
= 551 W

When a hot object of temperature TH radiates heat energy to its cool surroundings of temperature TC, the rate of net radiation heat loss can be expressed as

Q/t=ε × σ × (TH4-TC4 ) × AH

where AH is the area of the hot emitting body.

Example 9

A hot object at 127°C radiates heat energy at 2000 J/s to the surroundings, whose temperature is 27°C. The object has a coefficient of emissivity equal to 0.8. What is the area of the hot object in 2>? Take σ = 5.6703 × 10-8 W/m2K4.

Solution 9


TH = 127°C = 400 K = 4 × 102 K
TC = 27°C = 300 K = 3 × 102 K
Q/t = 2000 J/s = 2 × 103 J/s = 2 × 103 W
ε = 0.8
σ = 5.6703 × 10-8 W/m2K4
A = ?

Applying the equation

Q/t = ε × σ × (TH4 - TC4 ) × AH

we obtain after rearranging and substitutions:

AH = Q/t/ε × σ × (TH4 - TC4 )
= 2 × 103/0.8 × 5.6703 × 10 - 8 × [(4 × 102) - (3 × 102)]4
= 2 × 103/4.53624 × 10-8 × (256 - 81) × 108
= 2000/794
= 2.52 m2

d) Evaporation

Evaporation is sometimes regarded as the fourth method of heat transfer. By definition, evaporation is a method of heat transfer through heat removal from a hot body by means of change in state of a covering liquid.

When a liquid evaporates from a hot surface, it takes off heat energy from it. As a result, the surface gets colder. For example, when we sweat in summer it is because our body wants to get rid of the excessive heat it has accumulated due to the thermal processes occurred when chemical reactions take place in food we consume. Many chemical reactions are associated with heat release. Our body feels comfortable with a difference in temperature of 10-15°C with the surroundings. In this case, heat energy flows naturally from our body to the surrounding air. However, in hot days, this difference in temperature (otherwise known as temperature gradient) is much smaller, in some cases it even becomes negative (the surrounding air is hotter than the human body). As a result, the body does not cool down properly. At this point, the evaporation process comes to help, as the body produces sweat and therefore, the excessive heat is removed from our body through evaporation of sweat. This is because the most energetic particles of liquid (sweat) get some heat energy from our body to make them pass in gaseous state and therefore, they leave the body surface.

Refrigerators and air conditioners use liquids with low boiling points to produce cooling effect through evaporation.

The rate of heat flow through evaporation by a human body is calculated through the equation

Q/t = h × A × (ps - p0)

where h is the coefficient of heat transfer through evaporation, A is the area of the human body, ps is the pressure of water vapour near the skin and p0 is the pressure of water vapour in the environment air. The pressure of the water vapour near the surface of the skin depends on the humidity of the environment and the degree of sweating.

Example 10

The vapour pressure near the skin on a certain hot sunny day is 85 kPa while the air pressure is 100 kPa. Given that the body area of an adult is about 1.8 m2, calculate the heat lost due to evaporation in 2 minutes caused by the sweating process in the given conditions, if the coefficient of heat transfer through evaporation when the body is fully covered by sweat is 0.099 W/Pa × m2.

Solution 10


ps = 25 kPa = 85 000 Pa
p0 = 100 kPa = 100 000 Pa
A = 1.8 m2
h = 0.099 W/Pa × m2
t = 2 min = 120 s
Q = ?

From the equation

Q/t = h × A × (ps - p0)

we obtain after rearranging and substitutions:

Q = h × A × t × (ps - p0 )
= 0.099 × 1.8 × 120 × (85 000 - 100 000)
= -320 760 J

This is a big value, so the body needs immediately water and energy replacement through drinking and food.

The sign minus means the given amount of heat energy is leaving the body.

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