Physics Lesson 13.4.1 - Heat Transfer

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Welcome to our Physics lesson on Heat Transfer, this is the first lesson of our suite of physics lessons covering the topic of Calorimetry (Heat Transfer), you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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

Q_{released by the hot object} = Q_{gained by the cold object}

We can write for the heat released by the hottest object

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

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 × (t_f - 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 t_f 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):

Q_{released by the hot object} = Q_{gained by the cold object}

Q₁ = Q₂

m₁ × c₁ × ∆t₁ = m₂ × c₂ × ∆t₂

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

we obtain after substituting the known values:

20 × 1 × (80 - t_f ) = 30 × 0.58 × (t_f - 10)
20 × (80 - t_f ) = 17.4 × (t_f - 10)
1600 - 20 × t_f = 17.4 × t_f - 174
37.4 × t_f = 1774
t_f = 1774/37.4
=47.4°C

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.

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 - t_f )

Thus, the specific heat capacity of the hot object is

c₁ = Q₁/m₁ × (t₁ - t_f )
= 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 × (t_f-t₂ )

Therefore, the specific heat capacity of the cold object is

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

You have reached the end of Physics lesson 13.4.1 Heat Transfer. There are 3 lessons in this physics tutorial covering Calorimetry (Heat Transfer), you can access all the lessons from this tutorial below.

More Calorimetry (Heat Transfer) Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.4	Calorimetry (Heat Transfer)
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.4.1	Heat Transfer
13.4.2	What is a Calorimeter?
13.4.3	Methods of Heat Transfer

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