Class 11 Physics (Physics Part-II) Chapter 10 Thermal Properties Of Matter

Key Concept: Basic Concepts

d) A form of energy transfer between bodies

[Solution Description] Scientifically, heat is defined as a form of energy transfer between bodies due to a temperature difference.

Key Concept: Real-world Problem Solving, Conceptual Integration

b) It ensures structural preservation of the material

[Solution Description] Maintaining low temperatures in freeze-drying is essential because it ensures that the material remains frozen while water (or other solvents) sublimates directly from the solid state to the gaseous state without passing through the liquid phase. This prevents structural damage and preserves the integrity and quality of the original material. Additionally, low temperatures minimize chemical reactions and microbial growth, maintaining product stability and safety.

Key Concept: Application

b) Thermostats

[Solution Description] Bimetallic strip thermometers are commonly used in thermostats, where they help in regulating temperature by switching heating or cooling devices on or off.

Key Concept: Ideal Gas Law Application

b) 5.33 m³

[Solution Description]
According to the ideal gas law, $PV = \mu RT$; if the number of moles $\mu$ and pressure $P$ are constant, then $V_1/T_1 = V_2/T_2$. Thus, solving for $V_2$:
$V_2 = V_1 \times \frac{T_2}{T_1} = 4 \times \frac{400}{300} = 4 \times \frac{4}{3} = \frac{16}{3} \approx 5.33 \text{ m}^3$

Key Concept: Detailed Process

d) It warms up and rises.

[Solution Description] During a hot day, the sun heats the surface of the ocean. The air in contact with the warm surface becomes warmer and less dense than the surrounding cooler air. Due to its decreased density, this warmer air rises. As it rises, cooler air from the surroundings moves in to take its place, creating a convection current.

Key Concept: Resistance and Temperature

a) $343.16 \, \text{K}$
[Solution Description] To solve the problem, we use the linear relation between resistance RR and temperature TT, given by:

$R−R0R1−R0=T−T0T1−T0\frac{R – R_0}{R_1 – R_0} = \frac{T – T_0}{T_1 – T_0}$

Where:

$R0=80.0 ΩR_0 = 80.0 \, \Omega$ is the resistance at the triple point of water (273.16 K),

$R1=100.0 ΩR_1 = 100.0 \, \Omega$ is the resistance at the unknown temperature T1T_1,

$T0=273.16 KT_0 = 273.16 \, \text{K}$ is the triple point temperature.

Rearranging the equation to solve for T1T_1:

$T1=T0+(R1−R0)(T1−T0)(R1−R0)T_1 = T_0 + \frac{(R_1 – R_0)(T_1 – T_0)}{(R_1 – R_0)}$

Substitute known values:

$T1=273.16+100−80100−80=343.16 KT_1 = 273.16 + \frac{100 – 80}{100 – 80} = 343.16 \, \text{K}$

Thus, the unknown temperature is $343.16 K343.16,{K}$.

Key Concept: Thermal Stress

c) $5.5 \times 10^{7}$ N/m²

[Solution Description]
Thermal stress $\sigma$ can be calculated using the formula:
$\sigma = Y \cdot \alpha_l \cdot \Delta T$
Substitute the values:
$\sigma = 2 \times 10^{11} \times 1.1 \times 10^{-5} \times 25$
$\sigma = 5.5 \times 10^{7} \, \text{N/m}^2$
Thus, the thermal stress developed is $5.5 \times 10^{7} \, \text{N/m}^2$.

Key Concept: Temperature Conversion

c) 303.15 K

[Solution Description] To convert Celsius to Kelvin, use the formula:
$T_K = t_C + 273.15$
where $t_C$ is the Celsius temperature. Substituting the given value:
$T_K = 30 + 273.15 = 303.15$
Therefore, 30°C is equivalent to 303.15 K.

Key Concept: Advanced Graph Interpretation, Real-World Application

b) 24.1°C

[Solution Description]
We start by calculating the heat needed to melt the ice completely:
$Q_{\text{melt}} = 1 \times 3.33 \times 10^5 = 333000 \, \text{J}$
Calculate the heat released by cooling water from 40°C to the final temperature $T_f$:
$Q_{\text{cool}} = 5 \times 4186 \times (40 – T_f)$
At equilibrium, heat gained by ice equals heat lost by water:
$333000 = 5 \times 4186 \times (40 – T_f)$
$333000 = 20930 \times (40 – T_f)$
$333000 = 837200 – 20930T_f$
$20930T_f = 504200$
$T_f = \frac{504200}{20930} \approx 24.1^\circ C$
Therefore, the final temperature of the system is approximately $24.1^\circ C$.

Key Concept: Thermal Equilibrium in Phase Change, Pressure-Temperature Relationship, Latent Heat and Boiling

c) Latent heat released; temperature and pressure increase.

[Solution Description]

In a closed system, when steam condenses into water, latent heat is released. This release of energy raises the temperature of the surroundings if the heat is not transferred away effectively. The added heat can increase the pressure within the system as well due to increased temperature in the gaseous phase according to the ideal gas law.

Therefore, latent heat release leads to a rise in both temperature and pressure in the surroundings.

Key Concept: Concept Recall

c) The temperature remains constant.

[Solution Description] When water reaches its boiling point, it begins to convert from liquid to gas through a process called boiling. During this phase change, despite continuous heat transfer, the temperature remains constant until all the water has vaporized.

Key Concept: Temperature Definition

c) Both hotness and coldness

[Solution Description] Temperature is fundamentally defined as a measure of hotness or coldness. Hence, it indicates the degree of heat present in an object or substance.

Key Concept: Ideal Gas Law Application, Absolute Temperature Concept

c) 8 atm

[Solution Description]
By Gay-Lussac’s Law for a given mass and constant volume (${P_1}/{T_1} = {P_2}/{T_2}$):
Initial conditions are $P_1 = 2 \, \text{atm}$ and $T_1 = 100 \, \text{K}$. The final temperature $T_2 = 400 \, \text{K}$. Applying the equation:
$P_2 = P_1 \times \frac{T_2}{T_1} = 2 \, \text{atm} \times \frac{400 \, \text{K}}{100 \, \text{K}} = 8 \, \text{atm}$
Therefore, the new pressure is 8 atm.

Key Concept: Principle of Calorimetry

c) Heat gained equals heat lost

[Solution Description] A calorimeter operates based on the principle that the heat gained by the water or any substance inside it is equal to the heat lost by another object placed within, aligning with the conservation of energy concept.

Key Concept: Latent Heat Application, Thermodynamic Equilibrium

c) Latent heat facilitates vaporization, with liquid and gas in equilibrium.

[Solution Description] At 100°C and at standard atmospheric pressure, water reaches its boiling point where it transitions from liquid to gas. During this phase change, the temperature remains constant despite continuous energy input, as this energy is absorbed as latent heat required for vaporization. Thus, both liquid and gas phases exist in thermal equilibrium.

Key Concept: Area Expansion Calculation

c) 0.000225 m²

[Solution Description] Calculate the change in area using $\Delta A = 2 \alpha_l A_0 \Delta T$. Here, $A_0 = (0.5)^2 = 0.25\, \text{m}^2$, $\alpha_l = 1.5 \times 10^{-5}\, \text{K}^{-1}$, and $\Delta T = 30\, \text{°C}$. Thus, $\Delta A = 2 \times 1.5 \times 10^{-5} \times 0.25 \times 30 = 0.000225\, \text{m}^2$.

Key Concept: Basic Concept

c) 350 J

[Solution Description] According to the principle of heat exchange, the heat lost by one body is equal to the heat gained by another body. Therefore, if an object loses 350 J of heat, the other object gains 350 J.

Key Concept: Interval Calculation

c) 180 °F

[Solution Description] On the Fahrenheit scale, the ice point is at 32 °F and the steam point is at 212 °F. The number of intervals between these points can be calculated as follows:

$212 – 32 = 180$

There are 180 degrees between the ice point and steam point on the Fahrenheit scale.

Key Concept: Multi-Step Problem Solving, Real-World Problem Solving

d) 23.2°C

[Solution Description] Initially, we need to calculate the energy required to convert the ice at -5°C to water at 0°C:

$$Q_{ice\,to\,water} = m \cdot c_{ice} \cdot \Delta T = 0.5 \, \text{kg} \times 2100 \, \text{J/kg/K} \times 5 \, \text{K} = 5250 \, \text{J}$$

Next, calculate the energy to melt the ice:

$$Q_{melt} = m \cdot L_f = 0.5 \, \text{kg} \times 3.33 \times 10^5 \, \text{J/kg} = 166500 \, \text{J}$$

Then, calculate the energy released by steam condensing to water at 100°C:

$$Q_{condense} = m \cdot L_v = 0.1 \, \text{kg} \times 22.6 \times 10^5 \, \text{J/kg} = 226000 \, \text{J}$$

Calculate the energy released when steam cools from 120°C to 100°C:

$$Q_{cool\,steam} = m \cdot c_{steam} \cdot \Delta T = 0.1 \, \text{kg} \times 2010 \, \text{J/kg/K} \times 20 \, \text{K} = 4020 \, \text{J}$$

Summing energies, the total energy from the steam is:

$$Q_{total\,steam} = Q_{condense} + Q_{cool\,steam} = 226000 \, \text{J} + 4020 \, \text{J} = 230020 \, \text{J}$$

Since $$Q_{total\,steam} is greater than Q_{ice\,to\,water} + Q_{melt}$$, all the ice melts, and extra energy will further increase the temperature.

Remaining energy for heating water:

$$Q_{remaining} = 230020 \, \text{J} – (5250 \, \text{J} + 166500 \, \text{J}) = 58270 \, \text{J}$$

Final temperature rise for 600 g of water:

$$\Delta T = \frac{Q_{remaining}}{m \cdot c_{water}} = \frac{58270 \, \text{J}}{0.6 \, \text{kg} \times 4186 \, \text{J/kg/K}} \approx 23.21 \, \text{°C}$$

Final equilibrium temperature:

$$T_{final} = 0 \, \text{°C} + 23.21 \, \text{°C} \approx 23.2 \, \text{°C}$$

Key Concept: Specific Heat Capacity

a) 125 J/kg°C

[Solution Description] To find the specific heat capacity, use the formula $s = \frac{1}{m} \frac{\Delta Q}{\Delta T}$.
Given: $\Delta Q = 250 \text{ J}$, $m = 1 \text{ kg}$, and $\Delta T = 2 \text{ °C}$.
Substituting these values into the formula:
$s = \frac{1}{1} \times \frac{250}{2} = \frac{250}{2} = 125 \text{ J/kg°C}$
Therefore, the specific heat capacity is 125 J/kg°C.

Key Concept: Heat Transfer Calculation

c) 16 W

[Solution Description] The rate of heat transfer ($P$) through a material can be determined using Fourier’s law of thermal conduction:
$P = \frac{kA(T_1 – T_2)}{d}$
where $k$ is the thermal conductivity, $A$ is the area, $T_1 – T_2$ is the temperature difference, and $d$ is the thickness. In this case, $k = 0.04 \text{ W m}^{-1} \text{ K}^{-1}$, $A = 10 \text{ m}^2$, $(T_1 – T_2) = 20 \text{ °C}$, and $d = 0.5 \text{ m}$.
Substitute the given values:
$P = \frac{0.04 \times 10 \times 20}{0.5} = 16 \text{ W}$
The rate of heat transfer is 16 W.

Key Concept: Specific Heat Capacity

b) $50 J/(kg \cdot °C)$
[Solution Description]
To find the specific heat capacity ($c$), we use the formula:
$c = \frac{Q}{m \Delta T}$
where $Q = 500\,J$, $m = 2\,kg$, and $\Delta T = 5^\circ C$. Substituting these values,
$c = \frac{500}{2 \times 5} = \frac{500}{10} = 50 \, J/(kg \cdot ^\circ C)$
Therefore, the specific heat capacity is $50 \, J/(kg \cdot ^\circ C)$.

Key Concept: Pressure Influence, Conceptual Understanding

a) Increases both boiling point and latent heat

[Solution Description] When atmospheric pressure increases, more energy is needed for the molecules in the liquid to escape into the gas phase. Therefore, the boiling point of water increases. As the temperature at which water boils increases, the amount of heat required to convert it from liquid to gas (latent heat of vaporization) also tends to increase because additional energy is required to overcome the increased intermolecular forces holding the liquid together.

Key Concept: Interval Comparison

b) Celsius scale has larger intervals

[Solution Description]

The Fahrenheit scale has 180 intervals between the freezing point (32°F) and boiling point (212°F). The Celsius scale has 100 intervals between 0°C and 100°C. Therefore, the intervals on the Fahrenheit scale are smaller than those on the Celsius scale as there are more divisions over the same temperature range.

Key Concept: Application of Formula

c) 1500 J

[Solution Description] The heat required can be calculated using the formula: $\Delta Q = \mu C_v \Delta T$, where $\Delta Q$ is the heat added, $\mu$ is the number of moles, $C_v$ is the molar specific heat capacity at constant volume, and $\Delta T$ is the change in temperature.
First, we substitute the given values into the equation:
$\Delta Q = 4\ \text{mol} \times 25\ \text{J mol}^{-1} \text{ K}^{-1} \times 15\ \text{K}$
Next, perform the multiplication:
$\Delta Q = 4 \times 25 \times 15 = 1500\ \text{J}$
Thus, the heat required is 1500 J.

Key Concept: Real-world Calculation, Net Radiation Loss

b) 1200 W

[Solution Description]

To calculate the net rate of radiant energy loss, we use the formula:

$H = e\sigma A(T^4 – T_s^4)$

where $e = 0.8$, $\sigma = 5.67 \times 10^{-8} \, \text{W m}^{-2} \text{K}^{-4}$, $T = 500 \, \text{K}$, $T_s = 300 \, \text{K}$. The surface area $A$ of the cylinder is calculated using:

$A = 2\pi r(l + r) = 2\pi \times 0.05 \times (1.5 + 0.05) = 2\pi \times 0.05 \times 1.55$

Simplifying, we find:

$A = 0.487 \, \text{m}^2$

Plugging in the values into the equation for H:

$H = 0.8 \times 5.67 \times 10^{-8} \times 0.487 \times ((500)^4 – (300)^4)$

Calculating each term step by step:

$(500)^4 = 6.25 \times 10^{10}, \quad (300)^4 = 8.1 \times 10^9$

$(500)^4 – (300)^4 = 5.439 \times 10^{10}$

Now substituting back:

$H = 0.8 \times 5.67 \times 10^{-8} \times 0.487 \times 5.439 \times 10^{10}$

Performing the multiplication results in:

$H \approx 1198.96 \, \text{W}$

Thus, the net rate of radiant energy loss is approximately 1200 W.

Key Concept: Conceptual Application

c) Water has a higher specific heat capacity, moderating temperature changes.

[Solution Description]

Water has a high specific heat capacity compared to land, which means it can absorb more heat without a significant change in temperature. This property allows water bodies to moderate the temperature by absorbing heat during the day and releasing it at night, leading to milder temperatures in coastal regions.

Key Concept: Basic Definition

c) The temperature at which a solid turns into a liquid

[Solution Description] The melting point of a substance is defined as the temperature at which a solid turns into a liquid. It is an essential property that characterizes a substance’s phase transition from solid to liquid.

Key Concept: Steady State Concept

a) When the temperature of both substances is equal

[Solution Description]

The steady state in a calorimetry experiment is reached when there is no net exchange of heat between the substances involved. This means that the heat lost by one substance equals the heat gained by another. Hence, their temperatures become constant over time.

Key Concept: Complex Phase Diagram Interpretation, Thermodynamic Calculations

a) Solid forms

[Solution Description] To solve this problem, we must analyze the phase diagram and identify the regions where different phases exist. At the triple point, all three phases — solid, liquid, and gas — coexist. If the temperature is kept constant at 200 K and the pressure is increased from the triple point: – Initially, increasing the pressure from 10 Pa should push the system into the region where solids are stable because higher pressures generally favor the solid phase in many substances. Hence, increasing the pressure while keeping the temperature constant will result in transitioning from the vapor or liquid phase to the solid phase.

Thus, the correct answer is that it transitions to the solid phase.

Key Concept: Definition Recall

a) The temperature at which a liquid turns into a solid

[Solution Description] The freezing point is defined as the temperature at which a liquid turns into a solid. It is a characteristic property of the substance.

Key Concept: Resistance and Temperature

a) 120 $\Omega$
[Solution Description] The formula for resistance in relation to temperature is given by $R = R_o [1 + \alpha (T – T_o)]$. Here, $R_o = 100\,\Omega$, $\alpha = 0.004\,^{\circ}\text{C}^{-1}$, $T = 50^{\circ}\text{C}$, and $T_o = 0^{\circ}\text{C}$.
First, calculate the change in resistance:
$R = 100 [1 + 0.004 \times (50 – 0)]$
$R = 100 [1 + 0.004 \times 50]$
$R = 100 [1 + 0.2]$
$R = 100 \times 1.2 = 120\,\Omega$
The resistance at 50°C is therefore $120\,\Omega$.

Key Concept: Heat Flow Formula

a) $H = KA \frac{T_C – T_D}{L}$

[Solution Description] The formula for heat flow $H$ through a material is given by:
$H = KA \frac{T_C – T_D}{L}$
where $K$ is the thermal conductivity, $A$ is the area through which heat flows, $T_C$ and $T_D$ are temperatures, and $L$ is thickness.

Key Concept: Heat Transfer Rate

b) 20 W

[Solution Description]
Using the formula $H = KA \frac{T_C – T_D}{L}$ where $K = 50 \, \text{J} \, \text{s}^{-1} \, \text{m}^{-1} \, \text{K}^{-1}$, $A = 0.02 \, \text{m}^2$, $T_C – T_D = 40^\circ C$, and $L = 2 \, \text{m}$,
$H = 50 \times 0.02 \times \frac{40}{2} = 20 \, \text{W}$

Key Concept: Definition Recall

b) Latent Heat of Fusion

[Solution Description] The heat required to change a solid into a liquid at constant temperature is called the latent heat of fusion.

Key Concept: Experimental Setup

b) To measure the temperature change accurately

[Solution Description]

The calorimeter in the experimental setup for Newton’s Law of Cooling is used to measure the change in temperature as the body cools over time, allowing verification of the predicted cooling curve.

Key Concept: Emissivity Calculation

b) 0.63

[Solution Description]
Use the Stefan-Boltzmann Law for emissivity:
$H = Ae\sigma T^4$
Rearrange to solve for $e$:
$e = \frac{H}{A\sigma T^4}$
Substitute the given values:
$e = \frac{150}{0.1 \times 5.67 \times 10^{-8} \times 1200^4}$
Calculate $T^4$:
$1200^4 = 2.0736 \times 10^{12}$
Now calculate $e$:
$e = \frac{150}{0.1 \times 5.67 \times 10^{-8} \times 2.0736 \times 10^{12}} \approx 0.63$

Key Concept: Conductivity Calculation

b) 133.33 $J/s$

[Solution Description]

To calculate the rate of heat transfer, we use the formula:

$H = \frac{KA(T_C – T_D)}{L}$

where $K = 200 \, J s^{-1} m^{-1} K^{-1}$, $A = 0.02 \, m^2$, $T_C = 150^\circ C$, $T_D = 50^\circ C$, and $L = 3 \, m$.

Substituting these values into the formula:

$H = \frac{(200)(0.02)(150 – 50)}{3} = \frac{400}{3} = 133.33 \, J/s$

Therefore, the rate of heat transfer is 133.33 $J/s$.

Key Concept: Heat Transfer in Gases, Heat and Thermodynamics

c) The temperature decreases as no heat enters or leaves.

[Solution Description] In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas undergoing adiabatic expansion, the relation between temperature and volume is governed by:
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$
where $\gamma$ is the heat capacity ratio ($C_p/C_v$). Given $V_2 = 2V_1$, the equation becomes:
$T_2 = T_1 \left(\frac{V_1}{2V_1}\right)^{\gamma-1} = T_1 \left(\frac{1}{2}\right)^{\gamma-1}$
It shows that $T_2 < T_1$, indicating a decrease in temperature for an adiabatic process.

Key Concept: Comparative Analysis

b) Water

[Solution Description] To solve this problem, we need to consider the specific heat capacities of each substance. The substance with the highest specific heat capacity will undergo the slowest increase in temperature for the same amount of applied heat. Specific heat capacity is a measure of how much heat energy is needed to raise the temperature of a given mass of a substance by 1 Kelvin.

Let’s assume typical values for specific heat capacities:
– Aluminium: $900 \, \text{J/kg}\cdot\text{K}$
– Water: $4184 \, \text{J/kg}\cdot\text{K}$
– Lead: $128 \, \text{J/kg}\cdot\text{K}$
– Iron: $450 \, \text{J/kg}\cdot\text{K}$
Among these, water has the highest specific heat capacity, meaning it requires more heat to change its temperature compared to others.
Therefore, the correct answer is that water will have the slowest increase in temperature when equal heat is applied.

Key Concept: Calculating Linear Expansion

c) 0.003825 m

[Solution Description] The formula for linear expansion is $\Delta L = L_0 \cdot \alpha_l \cdot \Delta T$, where $L_0$ is the initial length, $\alpha_l$ is the coefficient of linear expansion and $\Delta T$ is the temperature change. Here, $L_0 = 3$ m, $\alpha_l = 1.7 \times 10^{-5}$ K$^{-1}$, and $\Delta T = 100 – 25 = 75$°C.
Calculating $\Delta L$:
$\Delta L = 3 \cdot 1.7 \times 10^{-5} \cdot 75 = 0.003825 \text{ m}$
So, the change in length is 0.003825 m.

Key Concept: Temperature Basics

c) The hotness or coldness of a body

[Solution Description] Temperature is a measure of the hotness or coldness of a body. It indicates how warm or cool an object is relative to another.

Key Concept: Calibration Process

c) It is universally constant

[Solution Description] The steam point or boiling point of water at standard atmospheric pressure is used as a fixed point because it is universally constant under standard conditions. This makes it reliable for calibrating thermometers across different locations and conditions.

Key Concept: Real-world Application, Pressure and Boiling

c) The boiling point decreases

[Solution Description] At lower pressures, such as those found at higher altitudes, the boiling point of water decreases. This is because the reduced external pressure requires less energy for the water molecules to escape into the gas phase. Therefore, at a high altitude with an atmospheric pressure of 60 kPa, the boiling point of water will be significantly lower than at standard atmospheric pressure (101.3 kPa).

Key Concept: Specific Heat Units

c) $J kg^{-1} K^{-1}$

[Solution Description] The SI unit of specific heat capacity is joules per kilogram per kelvin $(J kg^{-1} K^{-1})$, representing the amount of heat energy required to raise the temperature of one kilogram of a substance by one kelvin.

Key Concept: Sublimation Conditions, Phase Diagram

c) Atmospheric pressure around 1 atm
[Solution Description] Usually, dry ice sublimates directly from solid to gas at atmospheric pressures lower than the triple point. For CO$_2$, the sublimation occurs at standard atmospheric pressure (1 atm) and a temperature of -78.5°C.
In a phase diagram for CO$_2$, the line separating the solid phase from the gas phase (sublimation curve) indicates the conditions under which dry ice can sublimate. Therefore, the condition necessary for sublimation of dry ice at -78.5°C is:
Atmospheric pressure must be equal to or below the sublimation pressure, which is around 1 atm at -78.5°C.