Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What is Thermal Equilibrium?
Definition: The energy transferred from one body to another only because of a difference in temperature between them is called heat.
Discussion:
i) It is to be noted that in thermodynamics, temperature has been defined first. Then heat energy is defined based on the difference in temperature. Hence, a statement like, “Heat is the cause and temperature is the effect” is not applicable in thermodynamics. In fact, there is no real need for such a simplification,
ii) The accepted convention for the direction of flow of heat is that, it flows from a body at a higher temperature to a body at a lower temperature.
iii) The statement, ‘only because of a difference in temperature’, has been used in the definition of heat because difference in other physical properties, between two bodies may cause a flow of other forms of energy. For example, a difference in pressure between two bodies brought in contact causes the transfer of mechanical energy.
iv) While temperature is an intrinsic property of a body, ‘heat’ is not ‘Heat of a body’ is a meaningless concept. Heat energy manifests itself only when it is transferred from one body to another. Hence, heat is energy in transit. The statement, ‘temperature of a body is 20 °C ’, is meaningful. But ‘heat of a body is 200 cal ’ is meaningless. Instead ‘heat transferred from A to B is 200 cal ’ is a correct statement.
v) The amount of heat contained in a body can never be measured. What we measure in calorimetry is the amount of heat absorbed or liberated by a body—not the ‘heat content’ of the body.
Thermal equilibrium : When two or more bodies are in thermal contact and the temperature of everyone of them is the same, then no heat is exchanged among them. This state is called thermal equilibrium.
Numerical Examples
Example 1.
What is the temperature which has the same value in Celsius and in Fahrenheit scales?
Solution:
Let the temperature be x degree. As per the question C = F = x
From the relation for equivalence of temperature scales, we \(\frac{C}{5}\) = \(\frac{F-32}{9}\).
∴ \(\frac{x}{5}\) = \(\frac{x-32}{9}\) or, 9x = 5x – 160
or, 4x = -160 ∴ x = -40
∴ -40 °C = -40 °F
Example 2.
A thermometer has its lower fixed point and upper fixed point marked as 0.5 and 101 respectively. What is the reading on this thermometer at 30 °C.
Solution:
Let the reading be t degree.
∴ \(\frac{t-0.5}{101-0.5}\) = \(\frac{C}{100}\) or, \(\frac{t-0.5}{100.5}\) = \(\frac{30}{100}\)
or, 10t – 5 = 301.5 or, t = 30.65 degree
Hence, the faulty thermometer reads 30.65 degree.
Example 3.
A faulty thermometer reads -0.5 °C in melting ice and 99 °C in boding water at the pressure of 747 mm of Hg. What is the correct temperature when the faulty thermometer reads 45 °C ? Actual boiling point of water is 99 °C at 734 mm of Hg.
Solution:
Actual boiling point of water at the pressure of 760 mm of Hg is 100 °C.
Now, a decrease in pressure by (760 – 734) or 26 mm of Hg decreases the boiling point of water by (100 – 99) = 1 °C.
∴ The decrease in pressure by (760 – 747) or, 13 mm of Hg decreases the boiling point of water by \(\frac{1}{26}\) × 13 = 0.5 °C .
Hence, the boiling point of water at 747 mm of Hg should be (100 – 0.5) = 99.5 °C.
Let the Correct temperature be x°C, when the faulty ther-mometer reads 45 °C.
Hence, \(\frac{x}{99.5}\) = \(\frac{45-(-0.5)}{99-(-0.5)}\) = \(\frac{45.5}{99.5}\) ∴ x = 45.5 °C.
Example 4.
A centimetre scale is attached with a thermometer of uniform bore. The thermometer reads 7.3 cm in melting ice, 23.8 cm in boiling water and 3.5 cm in a freezing mixture. What is the temperature of this freezing mixture ¡n C?
Solution:
The lower and the upper fixed points correspond to readings of 7.3 cm and 23.8 cm respectively. The temperature of the freezing mixture in this scale corresponds to a scale reading of 3.5 cm.
Let C be the freezing mixture’s temperature in degree celsius.
∴ \(\frac{3.5-7.3}{23.8-7.3}\) = \(\frac{C-0}{100-0}\) or, C = \(\frac{-3.8}{16.5}\) × 100 = -23.03
∴ Temperature of the freezing mixture = -23.03 °c.
Alternative solution:
When the temperature increases from 0 °C to 100 °C, the corresponding change in scale reading = 23.8 – 7.3 = 16.5 cm. So when the temperature changes by 1 °C, the corresponding changes in scale reading = \(\frac{16.5}{100}\) = 0.165 cm
Let the temperature of the freezing mixture = -x °C
So change in temperature in Celcius scale = 0 – (-x) = x °C
Corresponding change in scale reading = 0.165x cm.
According to the question,
0.165x = 7.3 – 3.5 or, x = \(\frac{3.8}{0.165}\) = 23.03
So the temperature of the freezing mixture is -23.03 °C
Example 5.
A substance is heated from 30 °C to 75 °C. What is the change in its temperature on the Fahrenheit scale and on the Kelvin scale?
Solution:
Let a temperature be C on the Celsius scale, F on the Fahrenheit scale and T on the Kelvin scale. We can write,
F = \(\frac{9}{5}\)C + 32 …… (1)
and T = C + 273 …… (2)
Differentiating equation (1) we get,
ΔF = \(\frac{9}{5}\)ΔC
Here, ΔC = 75 – 30 = 45 ∴ ΔF = \(\frac{9}{5}\) × 45 = 81
Similarly, differentiating equation (2) we get,
ΔT = ΔC = 45
Example 6.
The graph between Ceiclus and Fahrenheit temperature of a body is shown in the Fig. Show that the angle made by the graph with Celsius axis is sin-1 \(\frac{9}{\sqrt{106}}\).
Solution:
We know,
\(\frac{C}{5}\) = \(\frac{F-32}{9}\) or, C = \(\frac{5}{9}\)(F – 32) or, F = \(\frac{9}{5}\)C + 32
This is the equation of a straight line,
Here the slope of the line, tan θ = \(\frac{9}{5}\)
