For an introduction to gyroscopes and an understanding of the gyroscopic effect and its applications, read this ScienceStruck article.
An Explanation of Thermal Equilibrium and its Formulas With Examples
This term might sound heavy and mysterious to the layman, but thermal equilibrium is an important condition in thermodynamics. This concept has wide variety of applications in nature and the human body too.
Thermodynamics is the branch of physics that deals with the relationship between heat and other properties (like pressure density temperature) in a substance. In the early 1800s, the use of steam engines led to the extensive study of heat and eventually the discovery of thermodynamics. The basis for most considerations in thermodynamics are the different laws of thermodynamics. These laws maintain that energy can be transferred from one system to the other only in the form of heat or work. During the initial study of thermodynamics scientists found that heat energy obtained from burning coal in steam engine could not be fully converted to mechanical work. Hence, they began studying about the energy contained in a system. They came up with a term named ‘entropy’ which could be quantified if the system (a liquid, gas or solid consisting of particles, whose motions determine its state) was at thermodynamic equilibrium.
Thermodynamic equilibrium of a system is said to be achieved when the system is in not only in thermal but also in mechanical, chemical as well as radioactive equilibrium. Applications of thermodynamics describes how systems respond to various changes in their surroundings. This can be applied to a wide variety of topics in science and engineering like engines, chemical reactions, phase transitions and even black holes.
What is Thermal Equilibrium?
When two substances having different temperatures are introduced or kept together, heat energy flows from a substance at higher temperature to a substance at lower temperature. Also, heat continues to be transferred till their temperatures are equalized. When this stage is achieved the substances are said to be in thermal equilibrium. Thus, we can generalize that ‘heat energy always flows from a hotter body to a colder body till their temperatures are equalized’. Or to put it differently – ‘A cold body does not become hot on its own, to raise its temperature, external heat energy must be supplied to it’.
We can also say, ‘the temperature of a substance determines the direction of heat flow between the substances, when they are introduced together’. So, equilibrium occurs when observable or macroscopic changes in a system cease to change with the passage of time. Thermal equilibrium of a system does not mean complete uniformity or stability within a system. For example, a river system can be in thermal equilibrium when the macroscopic or externally measurable temperature is stable and not changing in time, even though the surrounding or spatial temperature distribution reflects degradation of water quality leading to changes in temperature. Another example is when the thermometer is placed in contact with the patient’s body, If the body temperature of the patient and that of the mercury in the clinical thermometer have reached thermal equilibrium, then the temperature of the thermometer is the same as the body temperature, hence the reading of the thermometer shows the body temperature of the patient. Further illustration of this point is observed when food is put in the refrigerator, the heat from the food is transferred into the air of the refrigerator. The refrigeration process continues till thermal equilibrium is achieved and temperature of the food is equal to the temperature of the air in the refrigerator.
Calculation of Thermal Equilibrium
When a cold and hot object are placed together they will eventually come to an equilibrium. This resulting final temperature will be somewhere in between the two initial temperatures. This is apparent as the colder object will heat up while the hotter object will cool down. A classic example of this is thermal equilibrium of water filled with ice, wherein the ‘system’ is the warm room, glass, ice and water. The temperature of the ice will increase thereby decreasing the temperature of water. So, evidently the amount of heat lost by the hotter object will be given to the colder object.
Qhot to Qcold (Law of Conservation of Energy)
It may seem that Qlost = Qgained, but this is not completely correct.
(Or use the Law of Conservation of Energy: ∆E = 0(closed system) thereby we can deduce
the thermal equilibrium formula
Q = m x cp x ∆T
where Q = Heat Flow (Heat lost or Heat gained)
m = Mass of the substance
cp = Specific heat capacity
∆T = (Tf – Ti) = Difference in temperature
Numerical Problem 1
A 12 gram piece of aluminum (cp =.215 cal/g°C) is at 70°C. It is placed in a beaker that contains 35 grams of 15°C water. Find out at what temperature will they come to thermal equilibrium?
(In other words: “Find the final temperature of the mixture.”)
Qobject1 = – Qobject2 (negative on either side)
Qaluminum = – Qwater (makes the equation specific to this problem)
mA x cpA x ∆TA = -mwcpw∆Tw
mAcpA(Tf – TiA) = -mwcpw(Tf -Tiw) (putting in the equation for ∆T)
(12 g)(.215)( Tf – 70°C) = – 35(1)( Tf – 15°C) (putting in cpwateris 1 cal/g°C)
(2.58)( Tf – 70) = – 35( Tf – 15) (dropped units to make the algebra easier)
2.58 Tf – 180.6 = – 35Tf +525 (distributive property of multiplication; make sure the 525 is positive now [- times – = +])
37.58Tf = 705.6 (combining like terms)
Tf = 18.8°C (put units back at the end)
Here we can see that the difference 18.8°C, is between the two objects’ having initial temperatures of 70°C and 15°C. The colder object heated up and the hotter object cooled down.
Numerical Problem 2
A ball of aluminum (cp = 0.897 J/g°C) has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup containing 100 ml of water at a temperature of 15.0°C. What will be the final, equilibrium temperature of the ball and the water?
Heat lost by Aluminum = Heat gained by water.
‘T’ = final temp.
Al: 100 g x 0.897J/g/°C x (150 – T) ΔT.
= 13,455 – 89.7T
H2O: 100 g x 4.18J/g/°C x (T – 15) ΔT.
= 418T – 6,270.
13,455 – 89.7T = 418T – 6,270.
13,455 + 6,270 = 418T + 89.7T.
19,725 = 507.7T
T = 19,725 / 507.7 = 38.85°C final temp. (39°C)
Here the final temperature obtained after inserting an aluminum ball in water is 39°C. As the hotter aluminum ball cooled down thereby increasing the temperature of water.
These practical examples gave us a glimpse about the calculation of thermal equilibrium. Its applications and results are important in fields of chemistry, physics, aerospace engineering, mechanical engineering, bio-medical engineering and material sciences to name a few.