Thermal Equilibrium


This topic is part of the HSC Physics course under the section Thermodynamics.

HSC Physics Syllabus

  • explain the concept of thermal equilibrium (ACSPH022)
  • analyse the relationship between the change in temperature of an object and its specific heat capacity through the equation `Q = mc\Delta T` (ACSPH020)

Thermal Equilibrium in Thermodynamics Explained

What is Thermal Equilibrium?

Thermal equilibrium is achieved when two systems in physical contact with each other no longer exchange any heat, meaning they have the same temperature. It's essential to understand that this does not necessarily mean the systems have the same amount of internal energy, just that their temperatures are equal.


Thermal equilibrium


If System A is at a higher temperature than System B and they are brought into contact, heat will flow from A (the hotter system) to B (the cooler system) until they both reach the same temperature. Once this happens, they are in thermal equilibrium, and no more heat exchange occurs.

It is important to understand that the final temperature shared by the substances at thermal equilibrium is not always the average of their initial temperatures (shown in the diagram above). Whether this occurs depends on the mass of the substances and their specific heat capacities. Two substances will only reach a temperature that is the average of their starting temperatures if they have equal mass and specific heat capacities.

Imagine a hot cup of tea placed in a cooler room. Over time, the tea will cool down, and the room might become slightly warmer. This exchange of heat will continue until the temperature of the tea matches the room's temperature. At this point, they have reached thermal equilibrium. No more heat energy is transferred between the tea and its surroundings.

Zeroth Law of Thermodynamics

The Zeroth Law of Thermodynamics states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.


Zeroth law of thermodynamics


In simpler terms, if System A is in thermal equilibrium with System C, and System B is also in thermal equilibrium with System C, then System A and System B must be in thermal equilibrium with each other.

This law essentially establishes the concept of temperature. It is the foundation upon which the notion of temperature scale is built, enabling us to measure and compare the temperature of different objects. The name "zeroth law" was assigned because after the first, second, and third laws of thermodynamics were already established, it was realised that this fundamental postulate should precede the other laws, and thus was named the "zeroth" law.

Example 1 – Thermal Equilibrium Between Same Substances

Two identical metal blocks, Block A and Block B have initial temperatures of 30ºC and 50ºC respectively. Both blocks are placed in contact within an isolated system. What will be their final temperature when thermal equilibrium is reached? Assume the masses and specific heat capacities of both blocks are the same.



When two objects of different temperatures come into contact, the hotter object will lose heat to the cooler object until they both have the same temperature. The amount of heat lost by the hotter object will be equal to the amount of heat gained by the cooler object.

$$\Delta T_A = T - 30 \\$$

$$\Delta T_B = 50 - T$$


Using the heat equation:

$$Q = mc\Delta T \\$$


Heat gained by Block A = Heat lost by Block B:

$$m \times c \times \Delta T_A = m \times c \times (-\Delta T_B)$$


Substituting the values:

$$m \times c \times (T - 30) = m \times c \times (50 - T)$$


Solving for T:

$$T - 30 = 50 -T$$

$$2T = 80$$

$$T = 40^\circ \text{C}$$

Example 2 – Thermal Equilibrium Between Different Substances

Suppose you have 0.2 kg of water initially at 20ºC and a 0.3 kg aluminium block initially at 80ºC. The two are placed in contact within an isolated system. After some time, they achieve thermal equilibrium. What is their final temperature?

Use the following information for this question:

  • cwater = 4180 J kg–1 K–1
  • caluminium = 900 J kg–1 K–1



Using the heat equation

$$Q = mc\Delta T$$


Heat gained by the water = Heat lost by the metal:

$$m_{\text{water}} \times c_{\text{water}} \times (T - 20) = m_{\text{metal}} \times c_{\text{metal}} \times (80 - T)$$


Substituting the values and solving for T:

$$0.2 \times 4180 \times (T - 20) = 0.3 \times 900 \times (80 - T)$$

$$T = 34.6^\circ \text{C}$$


    RETURN TO MODULE 3: Waves and Thermodynamics