Welcome to Lecture 91 of our comprehensive JEE Physics Master Series. In this lecture, we commence Unit 8: Thermodynamics with the foundational bedrock of thermal science: Thermal Equilibrium, the Concept of Temperature, and the Zeroth Law of Thermodynamics. We formalize thermodynamic systems and boundaries, establish the physical definition of temperature through transitivity, examine the operational principles of thermometry, analyze the modern single-fixed-point Kelvin scale, and master the calibration of faulty and arbitrary temperature scales.
1. Detailed Concept Card
1.1 Thermodynamic Systems, Surroundings & Boundaries
In thermodynamics, we partition the universe into two distinct parts:
- System: A macroscopic portion of the universe chosen for thermodynamic investigation, containing a large number of particles ($N \sim 10^{23}$).
- Surroundings: Everything external to the system that can interact with it or exchange energy and matter.
- Boundary (Wall): The real or imaginary envelope separating the system from its surroundings. Boundaries are classified by their transmission properties:
- Diathermic Wall: A conducting boundary that permits free heat energy transfer across it ($\Delta Q \neq 0$), but prevents mass transfer (e.g., a thin copper plate).
- Adiabatic Wall: A thermally insulating boundary that strictly prevents heat energy transfer across it ($\Delta Q = 0$) (e.g., a thick vacuum-jacketed Dewar flask or asbestos sheet).
- Rigid vs Movable Wall: A rigid wall prevents mechanical work ($\Delta W = 0$), whereas a movable piston allows mechanical expansion/compression work ($\Delta W \neq 0$).
- Classification of Systems:
- Open System: Can exchange both energy and matter with surroundings (e.g., boiling water in an open beaker).
- Closed System: Can exchange energy (heat and work) but NOT matter with surroundings (e.g., gas sealed in a cylinder with a movable piston).
- Isolated System: Cannot exchange energy OR matter with surroundings (e.g., contents of an ideal vacuum flask).
1.2 Thermodynamic State Variables: Intensive vs Extensive
The macroscopic condition of a system is described by State Variables (thermodynamic coordinates):
- Intensive Variables: Independent of the total size, mass, or quantity of matter in the system. If a system in equilibrium is divided into two equal halves by an imaginary partition, intensive variables remain unchanged: $$\text{Pressure } (P), \quad \text{Temperature } (T), \quad \text{Density } (\rho), \quad \text{Specific Heat } (s), \quad \text{Refractive Index } (\mu)$$
- Extensive Variables: Directly proportional to the total mass or size of the system. In a partitioned system, extensive variables are halved: $$\text{Volume } (V), \quad \text{Mass } (M), \quad \text{Internal Energy } (U), \quad \text{Entropy } (S), \quad \text{Heat Capacity } (C)$$
- The Ratio Rule: The ratio of any two extensive properties is always an intensive property (e.g., $\rho = \frac{M}{V}$, $s = \frac{C}{M}$).
1.3 Thermal Equilibrium vs Complete Thermodynamic Equilibrium
A system is said to be in Complete Thermodynamic Equilibrium if and only if it satisfies three concurrent equilibria:
- 1. Mechanical Equilibrium: There are no unbalanced forces in the interior of the system or between the system and its surroundings ($F_{\text{net}} = 0$). Pressure $P$ is spatially uniform throughout.
- 2. Chemical Equilibrium: There is no net chemical reaction, no phase change, and no spontaneous diffusion of matter across concentration gradients. The chemical potential is uniform.
- 3. Thermal Equilibrium: When two systems separated by a diathermic boundary undergo no net exchange of heat energy, and their macroscopic thermodynamic coordinates cease to change with time, the systems are in Thermal Equilibrium.
1.4 The Zeroth Law of Thermodynamics & Concept of Temperature
Formulated formally by Ralph H. Fowler in 1935 (after the First and Second Laws were already established, hence designated “Zeroth”):
$$\mathbf{\text{If two systems } A \text{ and } B \text{ are each in thermal equilibrium with a third system } C,}$$
$$\mathbf{\text{then they are also in thermal equilibrium with each other.}}$$
Physical Significance of the Zeroth Law:
- Transitivity of Equilibrium: Thermal equilibrium is a mathematical equivalence relation (reflexive, symmetric, and transitive).
- Existence of Temperature ($T$): The Zeroth Law rigorously proves the existence of a unique, scalar state function called Temperature.
- Two systems are in thermal equilibrium if and only if their temperatures are numerically identical: $$T_A = T_C \quad \text{and} \quad T_B = T_C \implies T_A = T_B$$
- Thermometer Principle: The third system $C$ serves as a calibrated measuring instrument—a Thermometer. When brought into thermal contact with body $A$, it indicates a reading; when brought into contact with body $B$, if it indicates the same reading, $A$ and $B$ are guaranteed to be in thermal equilibrium without needing direct contact.
1.5 Thermometry & Thermometric Properties
To measure temperature quantitatively, one selects a physical property $X$ of a substance that varies continuously, reproducibly, and monotonically with temperature, called a Thermometric Property:
| Thermometer Type | Thermometric Substance | Thermometric Property ($X$) |
|---|---|---|
| Liquid-in-Glass (Mercury) | Mercury | Length of liquid column ($L$) |
| Constant-Volume Gas | Ideal gas ($\text{H}_2, \text{He}, \text{N}_2$) | Gas pressure ($P$) |
| Constant-Pressure Gas | Ideal gas | Gas volume ($V$) |
| Platinum Resistance (PRT) | Platinum wire | Electrical resistance ($R$) |
| Thermoelectric (Thermocouple) | Two dissimilar metals | Thermoelectric EMF ($\mathcal{E}$) |
| Optical Pyrometer | Thermal radiation | Spectral radiance / brightness |
A. Two-Fixed-Point Calibration Scale:
Historically, thermometer scales were defined by assigning arbitrary numbers to two reproducible fixed points under standard atmospheric pressure ($1.013 \times 10^5\text{ Pa}$):
- Lower Fixed Point (LFP, Ice Point): Melting ice at $0^\circ\text{C}$ ($32^\circ\text{F}$). Thermometric property value $= X_0$.
- Upper Fixed Point (UFP, Steam Point): Boiling water at $100^\circ\text{C}$ ($212^\circ\text{F}$). Thermometric property value $= X_{100}$.
Assuming linear variation $X(T) = a T + b$:
$$\mathbf{T = \left(\frac{X – X_0}{X_{100} – X_0}\right) \times 100^\circ\text{C}}$$
B. Modern Single-Fixed-Point Kelvin Scale (Triple Point of Water):
Since the 1954 CGPM convention, the international standard temperature scale utilizes a single reproducible fixed point: the Triple Point of Water ($T_{\text{tr}}$), where pure ice, liquid water, and water vapor coexist in unique three-phase thermodynamic equilibrium at $P_{\text{tr}} = 611.65\text{ Pa} \approx 4.58\text{ mm Hg}$:
$$\mathbf{T_{\text{tr}} \equiv 273.16\text{ K} \quad (0.01^\circ\text{C})}$$
For a thermometer with property $X$ (with $X = 0$ at $T = 0\text{ K}$):
$$\mathbf{T = 273.16 \left(\frac{X}{X_{\text{tr}}}\right)}$$
For a Constant-Volume Gas Thermometer in the limit of zero gas density ($P \to 0$ where all real gases behave identically as ideal gases):
$$\mathbf{T = 273.16 \lim_{P_{\text{tr}} \to 0} \left(\frac{P}{P_{\text{tr}}}\right)_V}$$
1.6 Inter-Conversion of Temperature Scales
Any arbitrary linear temperature scale $Y$ with Lower Fixed Point $\text{LFP}_Y$ and Upper Fixed Point $\text{UFP}_Y$ satisfies the universal invariant ratio:
$$\mathbf{\frac{\text{Reading} – \text{LFP}}{\text{UFP} – \text{LFP}} = \text{constant}}$$
Applying this fundamental relation across the classical temperature scales:
$$\mathbf{\frac{C – 0}{100} = \frac{F – 32}{180} = \frac{K – 273.15}{100} = \frac{R – 0}{80}}$$
Simplifying:
$$\mathbf{\frac{C}{5} = \frac{F – 32}{9} = \frac{K – 273.15}{5} = \frac{R}{4}}$$
Temperature Intervals:
A temperature change of $1^\circ\text{C}$ represents the exact same temperature increment as $1\text{ K}$:
$$\mathbf{\Delta C = \Delta K = \frac{5}{9}\Delta F}$$
1.7 Critical Conceptual Insights & JEE Traps
- Trap 1: “A body contains heat.” Heat is NEVER a state function or a stored property of a body. A body possesses Internal Energy ($U$) and has a measurable Temperature ($T$). Heat is exclusively energy in transit across a boundary driven by a temperature gradient.
- Trap 2: Triple Point ($273.16\text{ K}$) vs Ice Point ($273.15\text{ K}$): The ice point is the melting point of ice under $1\text{ atm}$ air pressure ($0.00^\circ\text{C} = 273.15\text{ K}$). The triple point is the equilibrium of pure water, ice, and vapor under its own vapor pressure without air ($0.01^\circ\text{C} = 273.16\text{ K}$). They differ by exactly $0.01\text{ K}$.
- Trap 3: Scale Invariance: Different thermometers calibrated at the same two fixed points ($0^\circ\text{C}$ and $100^\circ\text{C}$) may give slightly different readings at intermediate temperatures (e.g., $50^\circ\text{C}$) because their thermometric properties ($R(T), L(T), \mathcal{E}(T)$) do not follow identical non-linear functions. Only the ideal gas thermometer provides a universal thermodynamic temperature scale.
2. Four Graded Solved Examples
Example 1: Calibration & Error Analysis of a Faulty Thermometer (JEE Main Level)
Problem Statement: A faulty mercury thermometer has its lower fixed point (melting point of pure ice) marked as $4.0^\circ\text{C}$ and its upper fixed point (boiling point of water at $1\text{ atm}$) marked as $96.0^\circ\text{C}$.
- (a) If this thermometer reads $60.0^\circ\text{C}$ in a chemical solution, calculate the true temperature on the standard Celsius scale.
- (b) What will this faulty thermometer read when the true temperature of a body is $30.0^\circ\text{C}$?
- (c) At what temperature does this thermometer indicate the correct reading?
Step-by-Step Solution:
Step 1: Formulate the linear calibration equation:
Let $T_f$ be the reading on the faulty scale and $C$ be the correct Celsius reading:
$$\frac{T_f – \text{LFP}_f}{\text{UFP}_f – \text{LFP}_f} = \frac{C – 0}{100 – 0}$$
$$\frac{T_f – 4.0}{96.0 – 4.0} = \frac{C}{100} \implies \frac{T_f – 4.0}{92.0} = \frac{C}{100}$$
Step 2: Calculate true temperature when $T_f = 60.0^\circ\text{C}$:
$$\frac{60.0 – 4.0}{92.0} = \frac{C}{100} \implies \frac{56.0}{92.0} = \frac{C}{100}$$
$$C = \frac{56.0 \times 100}{92.0} = \frac{5600}{92} \approx \mathbf{60.87^\circ\text{C}}$$
Step 3: Calculate faulty reading when true temperature $C = 30.0^\circ\text{C}$:
$$\frac{T_f – 4.0}{92.0} = \frac{30.0}{100} = 0.30$$
$$T_f – 4.0 = 0.30 \times 92.0 = 27.60 \implies T_f = 27.60 + 4.0 = \mathbf{31.60^\circ\text{C}}$$
Step 4: Find temperature of identical reading ($T_f = C$):
$$\frac{C – 4.0}{92.0} = \frac{C}{100} \implies 100(C – 4.0) = 92 C$$
$$100 C – 400 = 92 C \implies 8 C = 400 \implies C = \mathbf{50.0^\circ\text{C}}$$
Key Takeaway: Below $50^\circ\text{C}$, the thermometer over-reads ($T_f > C$); above $50^\circ\text{C}$, it under-reads ($T_f < C$). At exactly $50^\circ\text{C}$, the two linear lines intersect and the error is zero.
Example 2: Platinum Resistance Thermometer (Electrical Resistance Scale)
Problem Statement: The electrical resistance of a platinum wire coil in a resistance thermometer is measured as $R_0 = 10.00\ \Omega$ at the ice point ($0^\circ\text{C}$) and $R_{100} = 13.80\ \Omega$ at the steam point ($100^\circ\text{C}$). When immersed in a hot thermostat oil bath, its resistance is measured as $R_t = 15.70\ \Omega$.
- (a) Calculate the temperature of the oil bath on the platinum resistance scale ($t_{\text{Pt}}$).
- (b) If the resistance of platinum follows the parabolic Callendar relation $R(t) = R_0(1 + A t + B t^2)$ where $A = 3.8 \times 10^{-3}\ ^\circ\text{C}^{-1}$ and $B = -5.8 \times 10^{-7}\ ^\circ\text{C}^{-2}$, explain why $t_{\text{Pt}}$ deviates from the true thermodynamic temperature $t$.
Step-by-Step Solution:
Step 1: Calculate temperature on the platinum scale:
By definition of the linear resistance temperature scale:
$$t_{\text{Pt}} = \left(\frac{R_t – R_0}{R_{100} – R_0}\right) \times 100^\circ\text{C}$$
Substitute numerical values:
$$R_t – R_0 = 15.70 – 10.00 = 5.70\ \Omega$$
$$R_{100} – R_0 = 13.80 – 10.00 = 3.80\ \Omega$$
$$t_{\text{Pt}} = \left(\frac{5.70}{3.80}\right) \times 100^\circ\text{C} = \frac{57}{38} \times 100 = 1.50 \times 100 = \mathbf{150.0^\circ\text{C}}$$
Step 2: Physical origin of scale divergence:
The platinum scale assumes that resistance increases strictly linearly with temperature ($R \propto t$). However, the true physical relation has a non-zero second-order term ($B t^2$). Because $B < 0$, the resistance curve bends slightly concave downward relative to the chord connecting $0^\circ\text{C}$ and $100^\circ\text{C}$, causing the linear platinum scale to predict a slightly higher temperature than the true thermodynamic temperature.
Key Takeaway: Two thermometers using different thermometric properties agree by definition at the fixed calibration points ($0^\circ\text{C}$ and $100^\circ\text{C}$), but depart at other temperatures due to property non-linearities.
Example 3: Constant-Volume Gas Thermometer & Absolute Kelvin Scale
Problem Statement: A constant-volume gas thermometer contains low-density helium gas. At the triple point of water ($T_{\text{tr}} = 273.16\text{ K}$), the gas pressure inside the bulb is recorded as $P_{\text{tr}} = 4.00 \times 10^4\text{ Pa}$.
- (a) When the bulb is immersed in a liquid boiling under standard atmospheric pressure, the pressure of the helium gas rises to $P = 5.20 \times 10^4\text{ Pa}$. Calculate the absolute thermodynamic temperature $T$ of the boiling liquid on the gas scale.
- (b) Convert this temperature to the Celsius scale ($^\circ\text{C}$).
- (c) What pressure would the thermometer indicate if cooled to absolute zero ($0\text{ K}$)?
Step-by-Step Solution:
Step 1: Calculate absolute temperature on the Kelvin scale:
For an ideal gas at constant volume, $P \propto T$. On the modern single-fixed-point scale:
$$T = 273.16 \times \left(\frac{P}{P_{\text{tr}}}\right)$$
Substitute numerical values:
$$T = 273.16 \times \left(\frac{5.20 \times 10^4\text{ Pa}}{4.00 \times 10^4\text{ Pa}}\right) = 273.16 \times 1.30 = \mathbf{355.11\text{ K}}$$
Step 2: Convert to Celsius:
$$C = T – 273.15 = 355.108 – 273.15 = \mathbf{81.96^\circ\text{C}}$$
Step 3: Gas pressure at absolute zero:
At $T = 0\text{ K}$, thermal kinetic energy of molecular motion ceases in classical thermodynamics, so:
$$P = P_{\text{tr}} \left(\frac{0}{273.16}\right) = \mathbf{0\text{ Pa}}$$
Key Takeaway: The constant-volume gas thermometer is the primary thermodynamic standard because in the limit $P \to 0$, all gases obey $PV = nRT$, making the measurement independent of the working gas identity.
Example 4: Scale Intersection Theorems & Arbitrary Scales (JEE Advanced Level)
Problem Statement:
- (a) Prove mathematically that the Celsius and Fahrenheit scales indicate the identical numerical reading at exactly $-40^\circ$.
- (b) Find the unique temperature at which the Fahrenheit and Kelvin scales register identical numerical values.
- (c) A novel temperature scale $W$ is devised where the freezing point of water is designated as $39^\circ\text{W}$ and the boiling point is $239^\circ\text{W}$. Formulate the linear conversion relation between $^\circ\text{W}$ and $^\circ\text{C}$, and find the reading on the $W$-scale corresponding to $30^\circ\text{C}$.
Step-by-Step Solution:
Step 1: Intersection of Celsius and Fahrenheit scales:
Let $C = F = x$ in the conversion formula:
$$\frac{x}{5} = \frac{x – 32}{9} \implies 9x = 5(x – 32) = 5x – 160$$
$$4x = -160 \implies x = \mathbf{-40^\circ} \implies \mathbf{-40^\circ\text{C} = -40^\circ\text{F}}$$
Step 2: Intersection of Fahrenheit and Kelvin scales:
Let $F = K = y$:
$$\frac{y – 32}{9} = \frac{y – 273.15}{5}$$
$$5(y – 32) = 9(y – 273.15) \implies 5y – 160 = 9y – 2458.35$$
$$4y = 2458.35 – 160 = 2298.35 \implies y = \frac{2298.35}{4} \approx \mathbf{574.59} \quad (\approx 574.25 \text{ if using } 273\text{ K})$$
$$\mathbf{574.59^\circ\text{F} = 574.59\text{ K}}$$
Step 3: Conversion relation for arbitrary scale $W$:
Lower fixed point $\text{LFP}_W = 39^\circ\text{W}$, Upper fixed point $\text{UFP}_W = 239^\circ\text{W}$:
$$\frac{W – \text{LFP}_W}{\text{UFP}_W – \text{LFP}_W} = \frac{C – 0}{100} \implies \frac{W – 39}{239 – 39} = \frac{C}{100}$$
$$\frac{W – 39}{200} = \frac{C}{100} \implies W – 39 = 2C \implies \mathbf{W = 2C + 39}$$
For $C = 30.0^\circ\text{C}$:
$$W = 2(30.0) + 39 = 60 + 39 = \mathbf{99^\circ\text{W}}$$
Key Takeaway: Any arbitrary scale is readily mapped to standard units by dividing the span into equal linear increments: $\frac{\text{Reading} – \text{LFP}}{\text{Fundamental Interval}} = \text{invariant}$.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. The Zeroth Law of Thermodynamics establishes the physical foundation for the definition of which of the following thermodynamic concepts?
- (A) Internal Energy
- (B) Temperature
- (C) Heat
- (D) Work
Q2. At what temperature do the Celsius and Fahrenheit temperature scales indicate the exact same numerical reading?
- (A) $+40^\circ$
- (B) $-40^\circ$
- (C) $0^\circ$
- (D) $-273.15^\circ$
Q3. A constant-volume gas thermometer registers a gas pressure of $P_0 = 80.0\text{ cm of Hg}$ at $0^\circ\text{C}$ and $P_{100} = 110.0\text{ cm of Hg}$ at $100^\circ\text{C}$. What pressure will it indicate when immersed in a bath maintained at $60^\circ\text{C}$?
- (A) $92.0\text{ cm of Hg}$
- (B) $95.0\text{ cm of Hg}$
- (C) $98.0\text{ cm of Hg}$
- (D) $100.0\text{ cm of Hg}$
Q4. On a newly designed thermometer scale $X$, the freezing point of water is marked as $-20^\circ\text{X}$ and the boiling point of water is marked as $180^\circ\text{X}$. If this thermometer reads $60^\circ\text{X}$, the corresponding temperature on the Celsius scale is:
- (A) $30^\circ\text{C}$
- (B) $40^\circ\text{C}$
- (C) $50^\circ\text{C}$
- (D) $60^\circ\text{C}$
Q5. In the modern international definition of the absolute temperature scale, the primary standard reference point is chosen as the triple point of water ($273.16\text{ K}$) rather than the normal melting point of ice because:
- (A) The triple point occurs at a higher temperature.
- (B) The triple point is a unique, fixed invariant state of nature independent of atmospheric pressure.
- (C) Ice melts at different temperatures under different gravitational fields.
- (D) The triple point can be measured only with a mercury thermometer.
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. Which of the following statements regarding thermodynamic systems, boundaries, and equilibrium states are correct?
- (A) Pressure and temperature are intensive state variables, whereas volume and internal energy are extensive state variables.
- (B) A diathermic wall allows free exchange of heat energy between two systems while preventing exchange of matter.
- (C) Complete thermodynamic equilibrium requires the simultaneous satisfaction of mechanical, chemical, and thermal equilibria.
- (D) If two isolated systems are brought into thermal contact through a rigid adiabatic wall, their temperatures will equalize spontaneously.
Q7. Which of the following statements regarding the Zeroth Law of Thermodynamics and thermometry are correct?
- (A) The Zeroth Law implies that thermal equilibrium is a transitive equivalence relation.
- (B) The Zeroth Law guarantees that temperature is a scalar property that determines whether systems are in thermal equilibrium.
- (C) In the limit of zero gas density ($P \to 0$), the constant-volume gas thermometer reading is completely independent of the identity of the gas.
- (D) Two different thermometers calibrated at the ice and steam points will always give identical readings at all intermediate temperatures regardless of the thermometric property chosen.
Q8. Which of the following statements regarding temperature scales and conversion intervals are correct?
- (A) A temperature change of $\Delta T = 10^\circ\text{C}$ corresponds exactly to a change of $10\text{ K}$.
- (B) A temperature change of $\Delta T = 10^\circ\text{C}$ corresponds exactly to a change of $18^\circ\text{F}$.
- (C) Absolute zero corresponds to $-273.15^\circ\text{C}$ and $-459.67^\circ\text{F}$.
- (D) The numerical reading on the Fahrenheit scale can never be negative.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A faulty thermometer has its lower fixed point marked at $5.0^\circ\text{C}$ and its upper fixed point marked at $95.0^\circ\text{C}$. When this thermometer measures the temperature of a patient and reads $59.0^\circ\text{C}$, calculate the true temperature of the patient in degrees Celsius ($^\circ\text{C}$).
Q10. The electrical resistance of a platinum wire is $5.00\ \Omega$ at $0^\circ\text{C}$ and $5.93\ \Omega$ at $100^\circ\text{C}$. When placed in a metallurgical furnace, the resistance of the wire increases to $7.79\ \Omega$. Determine the temperature of the furnace (in $^\circ\text{C}$) on the platinum resistance scale.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (B) | Zeroth Law & Concept of Temperature |
| Q2 | (B) | Scale Intersection Point |
| Q3 | (C) | Constant-Volume Gas Thermometer |
| Q4 | (B) | Arbitrary Scale Conversion |
| Q5 | (B) | Triple Point of Water Standard |
| Q6 | (A), (B), (C) | Thermodynamic Variables & Equilibria |
| Q7 | (A), (B), (C) | Zeroth Law Foundations & Thermometry |
| Q8 | (A), (B), (C) | Temperature Intervals & Absolute Zero |
| Q9 | 60 | Faulty Thermometer Correction |
| Q10 | 300 | Platinum Resistance Thermometry |
Step-by-Step Solutions
Solution 1:
The Zeroth Law of Thermodynamics establishes that thermal equilibrium is transitive ($A \sim C \text{ and } B \sim C \implies A \sim B$). This directly proves the existence of a scalar property called Temperature, whose equality governs thermal equilibrium. Correct Option: (B).
Solution 2:
Setting $C = F = x$ in the conversion formula:
$$\frac{x}{5} = \frac{x – 32}{9} \implies 9x = 5x – 160 \implies 4x = -160 \implies x = -40^\circ$$
Thus, $-40^\circ\text{C} = -40^\circ\text{F}$. Correct Option: (B).
Solution 3:
For a constant-volume gas thermometer with linear calibration:
$$T = \left(\frac{P – P_0}{P_{100} – P_0}\right) \times 100 \implies 60 = \left(\frac{P – 80}{110 – 80}\right) \times 100$$
$$\frac{P – 80}{30} = \frac{60}{100} = 0.60 \implies P – 80 = 0.60 \times 30 = 18.0$$
$$P = 80 + 18 = 98.0\text{ cm of Hg}$$
Correct Option: (C).
Solution 4:
Using the invariant calibration ratio:
$$\frac{X – \text{LFP}_X}{\text{UFP}_X – \text{LFP}_X} = \frac{C}{100} \implies \frac{60 – (-20)}{180 – (-20)} = \frac{C}{100}$$
$$\frac{80}{200} = \frac{C}{100} \implies 0.40 = \frac{C}{100} \implies C = 40^\circ\text{C}$$
Correct Option: (B).
Solution 5:
The normal boiling point and melting point of water depend strongly on local atmospheric pressure and ambient air dissolved in water. By contrast, the triple point of water occurs at a unique, fixed pressure ($611.65\text{ Pa}$) and temperature ($273.16\text{ K}$) determined strictly by nature without any external variable. Correct Option: (B).
Solution 6:
- (A) is Correct: Intensive variables are independent of system mass; extensive variables scale with mass.
- (B) is Correct: A diathermic wall is a thermal conductor permeable to heat but impermeable to matter.
- (C) is Correct: Thermodynamic equilibrium is complete only when mechanical, chemical, and thermal equilibria are simultaneously satisfied.
- (D) is Incorrect: An adiabatic wall prevents heat transfer; systems separated by an adiabatic wall cannot exchange heat and their temperatures will not equalize.
Correct Options: (A), (B), (C).
Solution 7:
- (A) is Correct: $A \sim C$ and $B \sim C \implies A \sim B$ is the definition of a transitive relation.
- (B) is Correct: The Zeroth Law guarantees that temperature is the common property equalized at thermal equilibrium.
- (C) is Correct: In the ideal gas limit ($P \to 0$), intermolecular forces vanish and all gases obey $PV = nRT$ identically.
- (D) is Incorrect: Different thermometric properties exhibit different non-linearities, causing thermometers to deviate at intermediate points between calibration marks.
Correct Options: (A), (B), (C).
Solution 8:
- (A) is Correct: $\Delta C = \Delta K$ because both scales have $100$ divisions between ice and steam points.
- (B) is Correct: $\Delta F = \frac{9}{5}\Delta C = \frac{9}{5}(10) = 18^\circ\text{F}$.
- (C) is Correct: Absolute zero ($0\text{ K}$) is $-273.15^\circ\text{C}$ and $\frac{9}{5}(-273.15) + 32 = -459.67^\circ\text{F}$.
- (D) is Incorrect: Temperatures below $-17.8^\circ\text{C}$ have negative Fahrenheit values down to $-459.67^\circ\text{F}$.
Correct Options: (A), (B), (C).
Solution 9:
Applying the faulty scale formula:
$$\frac{T_f – \text{LFP}}{\text{UFP} – \text{LFP}} = \frac{C}{100} \implies \frac{59.0 – 5.0}{95.0 – 5.0} = \frac{C}{100}$$
$$\frac{54.0}{90.0} = \frac{C}{100} \implies \frac{6}{10} = \frac{C}{100} \implies C = 0.60 \times 100 = 60^\circ\text{C}$$
Final Answer: 60
Solution 10:
On the platinum resistance scale:
$$t = \left(\frac{R_t – R_0}{R_{100} – R_0}\right) \times 100^\circ\text{C}$$
$$R_t – R_0 = 7.79 – 5.00 = 2.79\ \Omega$$
$$R_{100} – R_0 = 5.93 – 5.00 = 0.93\ \Omega$$
$$t = \left(\frac{2.79}{0.93}\right) \times 100 = 3.00 \times 100 = 300^\circ\text{C}$$
Final Answer: 300
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