Thermal Properties Of MatterClass 11 Physics NCERT Solutions

Concept: The temperature on any absolute scale is proportional to the corresponding temperature on the Kelvin scale. Let a temperature be $T_K$ on the Kelvin scale, $T_A$ on scale A, and $T_B$ on scale B.

Then, $\frac{T_K}{273.16} = \frac{T_A}{200} = \frac{T_B}{350}$

Calculation: To find the relation between $T_A$ and $T_B$, we can use the following part of the above relation:

Now, we solve for $T_A$ in terms of $T_B$:

Final Answer: The relation between the temperatures on the two scales is $T_A = \frac{4}{7} T_B$.

Concept: The triple point of water is a fixed physical point and has an absolute temperature of $273.16$ K on the Kelvin scale. We can establish a relationship between scales A, B, and Kelvin.

Calculation: From the given information: $200 \text{ A} = 273.16 \text{ K}$ $1 \text{ A} = \frac{273.16}{200} \text{ K}$

Also, $350 \text{ B} = 273.16 \text{ K}$ $1 \text{ B} = \frac{273.16}{350} \text{ K}$

Let $T_A$ and $T_B$ be the temperatures of a body on scales A and B respectively. The corresponding temperature in Kelvin, $T_K$, would be: $T_K = T_A \times (\frac{273.16}{200})$ $T_K = T_B \times (\frac{273.16}{350})$

Equating the two expressions for $T_K$: $T_A \times \frac{273.16}{200} = T_B \times \frac{273.16}{350}$ Dividing both sides by 273.16: $\frac{T_A}{200} = \frac{T_B}{350}$ Simplifying the relation: $\frac{T_A}{T_B} = \frac{200}{350} = \frac{20}{35} = \frac{4}{7}$ So, $7T_A = 4T_B$ or $T_A = \frac{4}{7} T_B$.

Final Answer: The relation between $T_A$ and $T_B$ is $\frac{T_A}{200} = \frac{T_B}{350}$ or $T_A = \frac{4}{7} T_B$.

Concept: Since the resistance varies linearly with temperature, we can assume that the ratio of the difference in temperature to the difference in resistance is constant. We can set up a proportion:

Calculation: Substitute the given values into the equation:

Now, solve for $T$:

Final Answer: The temperature when the resistance is $123.4 \Omega$ is approximately $384.82$ K.

Step 1: Find the temperature coefficient of resistance, $\alpha$. Using the data for the melting point of lead: $R_1 = R_o[1 + \alpha(T_1 - T_o)]$ $165.5 = 101.6[1 + \alpha(600.5 - 273.16)]$ $\frac{165.5}{101.6} = 1 + \alpha(327.34)$ $1.6289 = 1 + 327.34\alpha$ $0.6289 = 327.34\alpha$ $\alpha = \frac{0.6289}{327.34} = 1.921 \times 10^{-3} \text{ K}^{-1}$

Step 2: Find the unknown temperature $T_2$. Using the given relation for the resistance $R_2$: $R_2 = R_o[1 + \alpha(T_2 - T_o)]$ $123.4 = 101.6[1 + (1.921 \times 10^{-3})(T_2 - 273.16)]$ $\frac{123.4}{101.6} = 1 + (1.921 \times 10^{-3})(T_2 - 273.16)$ $1.2146 = 1 + (1.921 \times 10^{-3})(T_2 - 273.16)$ $0.2146 = (1.921 \times 10^{-3})(T_2 - 273.16)$ $T_2 - 273.16 = \frac{0.2146}{1.921 \times 10^{-3}}$ $T_2 - 273.16 = 111.71$ $T_2 = 111.71 + 273.16 = 384.87 \text{ K}$

Final Answer: The temperature when the resistance is $123.4 \Omega$ is approximately $384.87$ K.

(a) Triple-point of water as a standard fixed point: The triple point of water is the unique condition of temperature and pressure at which water, ice, and water vapor can coexist in stable equilibrium. This occurs at a precise temperature of 273.16 K and a pressure of 611.657 Pa. Because it is a single, unique, and easily reproducible point, it serves as an excellent standard fixed point.

The problem with using the melting point of ice and the boiling point of water is that both of these points vary with pressure. For example, the boiling point of water decreases at higher altitudes where the atmospheric pressure is lower. Therefore, they are not unique points unless the pressure is also strictly specified (e.g., at standard atmospheric pressure), making them less reliable as fundamental standards compared to the triple point.

(b) The other fixed point on the Kelvin scale: The Kelvin scale is an absolute temperature scale. It has only one standard fixed point, which is the triple point of water (273.16 K). The other fixed point is the absolute zero (0 K), which is the theoretical temperature at which all molecular motion ceases. This is the natural zero of the scale, not an arbitrarily chosen point.

(c) The use of 273.15 in the conversion formula: The relation $t_c = T - 273.15$ defines the Celsius scale in terms of the Kelvin scale. On the Celsius scale, the freezing point of water at standard atmospheric pressure is defined as $0^{\circ}$C. This temperature corresponds to 273.15 K on the Kelvin scale.

The triple point of water is at 273.16 K. This corresponds to $t_c = 273.16 - 273.15 = 0.01^{\circ}$C. The number 273.15 is used in the conversion to ensure that the zero of the Celsius scale ($0^{\circ}$C) remains the melting point of ice at 1 atm pressure, which is a historically convenient reference point.

(d) Triple-point of water on a Fahrenheit-sized absolute scale: The size of a unit interval on the Fahrenheit scale is different from that on the Celsius or Kelvin scale. The relationship is: 100 Celsius degrees = 180 Fahrenheit degrees. So, 1 Celsius degree = $\frac{180}{100} = \frac{9}{5}$ Fahrenheit degrees. Since the Kelvin scale has the same unit interval size as the Celsius scale, 1 K = $\frac{9}{5}$ units on the new absolute scale.

The triple point of water is at $T = 273.16$ K. To express this on an absolute scale with unit interval size equal to Fahrenheit, we multiply the Kelvin temperature by $\frac{9}{5}$.

Temperature on new scale = $273.16 \times \frac{9}{5} = 491.688$

This scale is known as the Rankine scale ($^{\circ}$R).

Final Answer: The temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale is 491.69.

(a) The triple point of water is chosen as a standard fixed point because it is a unique and reproducible state. It occurs at a specific temperature ($273.16$ K) and pressure ($611.73$ Pa) where ice, liquid water, and water vapor coexist in equilibrium. This condition is independent of external factors, making it a highly precise reference. In contrast, the melting point of ice and the boiling point of water both depend on the surrounding pressure. For them to be used as fixed points, the pressure must be standardized to 1 atmosphere, which is more difficult to reproduce with high accuracy than the conditions for the triple point.

(b) On the Kelvin absolute scale, there is only one standard fixed point, which is the triple point of water (defined as 273.16 K). The other reference point is absolute zero (0 K), which is the theoretical temperature at which all molecular motion ceases. The scale is defined by setting the temperature of absolute zero to 0 K and the triple point of water to 273.16 K.

(c) The relation $t_c = T - 273.15$ uses 273.15 because the zero of the Celsius scale ($0^\circ\text{C}$) is defined as the freezing point of water at standard atmospheric pressure, not the triple point. The triple point of water, defined as 273.16 K, occurs at a slightly higher temperature than the normal freezing point. The freezing point of water at 1 atm is approximately 273.15 K. Therefore, to align the zero of the Celsius scale with the Kelvin scale, the offset is 273.15.

(d) An absolute scale with a unit interval size equal to the Fahrenheit scale is the Rankine scale ($^\circ\text{R}$). The relationship between Kelvin (K) and Fahrenheit ($^\circ\text{F}$) interval sizes is: $100 \text{ K} = 180 ^\circ\text{F}$, or $1 \text{ K} = \frac{9}{5} ^\circ\text{F}$. The Rankine scale is related to the Kelvin scale by $T_R = \frac{9}{5} T_K$. The triple point of water is defined as $T_K = 273.16$ K. Therefore, on this absolute scale (Rankine scale), the temperature of the triple point of water is: $T_R = \frac{9}{5} \times 273.16 = 491.688 \text{ R}$

Final Answer: The temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale is $491.69$ R.

For an ideal gas thermometer, the absolute temperature $T$ is directly proportional to the pressure $P$ at constant volume ($T \propto P$). Therefore, we can write: $\frac{T}{T_{tr}} = \frac{P}{P_{tr}}$ where $T_{tr}$ is the temperature of the triple point of water (273.16 K), and $P_{tr}$ is the pressure at that temperature. Thus, the unknown temperature $T$ (melting point of sulphur) can be calculated as: $T = T_{tr} \times \frac{P}{P_{tr}} = 273.16 \times \frac{P}{P_{tr}}$

Reason for difference: The assumption that the gases used (oxygen and hydrogen) are perfectly ideal is not completely accurate. Real gases deviate from ideal gas behavior, especially at non-zero pressures. The intermolecular forces and finite size of molecules cause this deviation. Hydrogen is closer to an ideal gas than oxygen, so the reading from thermometer B is likely more accurate.

Procedure to reduce discrepancy: To get a more accurate value of the temperature, the measurements should be repeated with smaller and smaller amounts of gas in the thermometer, which corresponds to taking readings at lower and lower pressures ($P_{tr}$ and $P$). A graph of the calculated temperature $T$ versus the pressure $P_{tr}$ should be plotted. The graph is then extrapolated to zero pressure ($P_{tr} \rightarrow 0$). The temperature value at this intercept corresponds to the behavior of a truly ideal gas and will be the same regardless of the gas used.

(a) Calculation of Sulphur's Melting Point: For an ideal gas thermometer operating at constant volume, pressure is directly proportional to the absolute temperature ($P \propto T$). Therefore, we can use the relation: $T = T_{tr} \times \frac{P}{P_{tr}}$ where $T_{tr} = 273.16$ K is the triple point temperature of water, and $P_{tr}$ is the pressure at that temperature.

(b) Reason for Discrepancy and Improvement: The slight difference in the readings is because the gases used, oxygen and hydrogen, are real gases and do not behave perfectly as ideal gases. Their behavior deviates from the ideal gas law, especially at the given pressures. Hydrogen, being a lighter molecule, behaves more like an ideal gas than oxygen, so its reading is likely closer to the true value.

To reduce the discrepancy and obtain a more accurate value, the experiment should be repeated with progressively lower pressures of the gases in the thermometers. The calculated temperatures for the melting point of sulphur should be plotted against the pressure at the triple point ($P_{tr}$). The graph for each gas should then be extrapolated to zero pressure ($P_{tr} \rightarrow 0$). At zero pressure, all real gases behave as ideal gases, and the extrapolated values from both thermometers should converge to the same, more accurate temperature.

Part 1: Actual length of the rod at $45.0^\circ\text{C}$. On the hot day, the steel tape expands. The markings on the tape become farther apart. Therefore, a length measured by the tape will be an underestimate of the true length. The actual length of any segment of the tape at temperature $T$ is given by $L' = L(1 + \alpha \Delta T)$, where $L$ is the length at the calibration temperature $T_0$ and $\Delta T = T - T_0$.

Change in temperature, $\Delta T = 45.0^\circ\text{C} - 27.0^\circ\text{C} = 18.0^\circ\text{C} = 18.0$ K.

The reading of 63.0 cm on the tape corresponds to an actual length that has also expanded. The actual length of the rod at $45.0^\circ\text{C}$ is: $L_{actual} = L_{measured} (1 + \alpha \Delta T)$ $L_{actual} = 63.0 \text{ cm} \times [1 + (1.20 \times 10^{-5} \text{ K}^{-1}) \times (18.0 \text{ K})]$ $L_{actual} = 63.0 \times [1 + 0.000216]$ $L_{actual} = 63.0 \times 1.000216 = 63.013608 \text{ cm}$

Part 2: Length of the rod at $27.0^\circ\text{C}$. Both the rod and the tape are made of steel and thus have the same coefficient of linear expansion. When a measurement is taken where both objects are at the same temperature, the expansion effects cancel out. Let $L_{rod, 27}$ be the length of the rod at $27^\circ\text{C}$. At $45^\circ\text{C}$, the true length of the rod is $L_{rod, 45} = L_{rod, 27}(1 + \alpha \Delta T)$. The true length corresponding to the tape reading of 63.0 cm at $45^\circ\text{C}$ is $L_{tape, 45} = 63.0(1 + \alpha \Delta T)$. Since the rod's length is measured by the tape, $L_{rod, 45} = L_{tape, 45}$. $L_{rod, 27}(1 + \alpha \Delta T) = 63.0(1 + \alpha \Delta T)$ $L_{rod, 27} = 63.0 \text{ cm}$ This means the length of the rod at the calibration temperature is exactly what the tape reads, regardless of the temperature at which the measurement is taken (as long as both rod and tape are at the same temperature).

On a hot day, the steel tape expands. Each 'cm' marking on the tape is now longer than a true cm. Therefore, the tape will under-read the actual length.

The actual length ($L_{actual}$) is given by the formula: $L_{actual} = L_{measured} [1 + \alpha(T - T_0)]$

$L_{actual} = 63.0 \text{ cm} \times [1 + (1.20 \times 10^{-5} \text{ K}^{-1}) \times (18.0 \text{ K})]$ $L_{actual} = 63.0 \times [1 + 0.000216]$ $L_{actual} = 63.0 \times 1.000216$ $L_{actual} = 63.013608 \text{ cm}$

So, the actual length of the steel rod at $45.0^{\circ}$C is approximately 63.0136 cm.

We now know the actual length of the rod at $45.0^{\circ}$C is $L_{45} = 63.0136$ cm. We need to find its length at $27.0^{\circ}$C ($L_{27}$). The rod will contract when cooled.

The formula for thermal expansion is $L_T = L_0(1 + \alpha \Delta T)$. Here, $L_{45}$ is the final length and $L_{27}$ is the initial length.

$L_{45} = L_{27} [1 + \alpha(45.0 - 27.0)]$ $63.0136 \text{ cm} = L_{27} [1 + (1.20 \times 10^{-5}) \times 18]$ $63.0136 = L_{27} [1.000216]$ $L_{27} = \frac{63.0136}{1.000216} = 63.0 \text{ cm}$

Alternatively, since the tape is calibrated for $27.0^{\circ}$C, the measurement of 63.0 cm at $45.0^{\circ}$C implies that the rod would have a length of exactly 63.0 cm at the calibration temperature of $27.0^{\circ}$C.

Concept: For the wheel to slip on the shaft, the diameter of the shaft must be equal to or slightly less than the diameter of the hole. We will calculate the temperature at which the shaft's diameter becomes exactly 8.69 cm.

The change in diameter due to a change in temperature is given by the formula for linear expansion: $D_2 = D_1 [1 + \alpha(T_2 - T_1)]$

Calculation: Substitute the given values into the formula:

Final Answer: The shaft must be cooled to a temperature of approximately $-69^{\circ}$C for the wheel to slip on it.

Concept: The wheel will slip onto the shaft when the diameter of the shaft contracts to become equal to the diameter of the hole. We use the formula for linear thermal expansion (or contraction).

Formula: $D_2 = D_1(1 + \alpha \Delta T)$ where $D_2$ is the final diameter, $D_1$ is the initial diameter, and $\Delta T = T_2 - T_1$ is the change in temperature.

Calculation: We need the final diameter of the shaft $D_2$ to be equal to the diameter of the hole, $d$. So, $D_2 = 8.69$ cm.

Substitute the values into the formula: $8.69 = 8.70[1 + (1.20 \times 10^{-5})(T_2 - 27)]$ $\frac{8.69}{8.70} = 1 + (1.20 \times 10^{-5})(T_2 - 27)$ $0.99885 = 1 + (1.20 \times 10^{-5})(T_2 - 27)$ $0.99885 - 1 = (1.20 \times 10^{-5})(T_2 - 27)$ $-0.00115 = (1.20 \times 10^{-5})(T_2 - 27)$ $T_2 - 27 = \frac{-0.00115}{1.20 \times 10^{-5}}$ $T_2 - 27 = -95.83$ $T_2 = -95.83 + 27 = -68.83^\circ\text{C}$

Final Answer: The wheel will slip on the shaft when the temperature of the shaft is cooled to approximately $-69^\circ\text{C}$.

Concept: When a sheet with a hole is heated, the hole expands in the same way as a solid disc of the same material would. The change in diameter can be calculated using the formula for linear thermal expansion.

Formula: $\Delta D = D_1 \alpha \Delta T$ where $\Delta T = T_2 - T_1$ is the change in temperature.

Calculation: First, calculate the change in temperature: $\Delta T = T_2 - T_1 = 227^\circ\text{C} - 27.0^\circ\text{C} = 200^\circ\text{C} = 200 \text{ K}$

Now, calculate the change in diameter: $\Delta D = (4.24 \text{ cm}) \times (1.70 \times 10^{-5} \text{ K}^{-1}) \times (200 \text{ K})$ $\Delta D = 4.24 \times 1.70 \times 10^{-5} \times 200 \text{ cm}$ $\Delta D = 4.24 \times 3.40 \times 10^{-3} \text{ cm}$ $\Delta D = 0.014416 \text{ cm}$

Final Answer: The change in the diameter of the hole is approximately 0.0144 cm.

Concept: When a sheet with a hole is heated, the hole expands in the same way as a solid disk of the same material would. The change in diameter can be calculated using the formula for linear expansion: $\Delta D = D_1 \alpha \Delta T$

Calculation: First, calculate the change in temperature, $\Delta T$: $\Delta T = T_2 - T_1 = 227 - 27 = 200^{\circ}\text{C} = 200 \text{ K}$

Now, substitute the values into the formula for the change in diameter:

Final Answer: The change in the diameter of the hole is approximately 0.0144 cm.

Concept: When the wire is cooled, it attempts to contract. Since it is held by rigid supports, this contraction is prevented, leading to the development of thermal stress and tension. The thermal strain is equal to the fractional change in length that would have occurred if the wire were free.

Calculation: Step 1: Calculate the change in temperature. $\Delta T = T_2 - T_1 = -39^\circ\text{C} - 27^\circ\text{C} = -66^\circ\text{C}$ $|\Delta T| = 66 \text{ K}$

Step 2: Calculate the thermal strain. $\text{Strain} = \alpha |\Delta T| = (2.0 \times 10^{-5} \text{ K}^{-1}) \times (66 \text{ K}) = 1.32 \times 10^{-3}$

Step 3: Calculate the cross-sectional area of the wire. Radius, $r = d/2 = 1.0$ mm = $1.0 \times 10^{-3}$ m. $A = \pi r^2 = \pi (1.0 \times 10^{-3} \text{ m})^2 = 3.1416 \times 10^{-6} \text{ m}^2$

Step 4: Calculate the tension developed. $F = Y \times A \times \text{Strain}$ $F = (0.91 \times 10^{11} \text{ Pa}) \times (3.1416 \times 10^{-6} \text{ m}^2) \times (1.32 \times 10^{-3})$ $F = (0.91 \times 3.1416 \times 1.32) \times 10^{11-6-3} \text{ N}$ $F = 3.77 \times 10^2 \text{ N} = 377 \text{ N}$

Final Answer: The tension developed in the wire is approximately 377 N.

Concept: When the wire is cooled, it tries to contract. Since it is held between rigid supports, this contraction is prevented, which develops a tensile stress in the wire. The tension is the force corresponding to this stress.

Final Answer: The tension developed in the wire is approximately 377 N.

Concept: The total change in length of the combined rod is the sum of the individual changes in length of the brass and steel rods. $\Delta L_{total} = \Delta L_{brass} + \Delta L_{steel}$ The change in length for each rod is given by $\Delta L = L_0 \alpha \Delta T$.

Since the ends of the combined rod are free to expand, each rod expands freely according to its own coefficient of expansion. There are no external forces restricting this expansion. Therefore, no thermal stress is developed at the junction or anywhere else in the rod.

Thermal stress only develops when the natural expansion or contraction of a material is constrained or prevented.

Part 1: Total change in length. First, calculate the change in temperature: $\Delta T = T_2 - T_1 = 250^\circ\text{C} - 40.0^\circ\text{C} = 210^\circ\text{C} = 210 \text{ K}$

The total change in length is the sum of the changes in length of the individual rods. Formula for change in length: $\Delta L = L \alpha \Delta T$.

Part 2: Thermal stress. Since the ends of the combined rod are free to expand, each rod expands freely according to its properties. There is no external force preventing this expansion. Therefore, no stress is developed at the junction or anywhere else in the rod.