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Step-by-step solutions with formulas and diagrams for all questions from NCERT Class 11 Physics Chapter 10.
Start Solving →Q1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Temperature can be expressed in different scales such as Kelvin (K), Celsius (°C), and Fahrenheit (°F). The Kelvin scale is an absolute thermodynamic scale where 0 K corresponds to absolute zero.
The conversion relations between temperature scales are:
$$ t_C = T_K - 273.15 $$ $$ t_F = \frac{9}{5}t_C + 32 $$where
Given:
$$ T = 24.57\ \text{K} $$Convert Kelvin → Celsius
$$ \begin{aligned} t_C &= T - 273.15 \\ &= 24.57 - 273.15 \\ &= -248.58^\circ\text{C} \end{aligned} $$Convert Celsius → Fahrenheit
$$ \begin{aligned} t_F &= \frac{9}{5}t_C + 32 \\ &= \frac{9}{5}(-248.58) + 32 \\ &= -447.444 + 32 \\ &= -415.44^\circ\text{F} \end{aligned} $$Thus, the triple point of neon is:
Given:
$$ T = 216.55\ \text{K} $$Convert Kelvin → Celsius
$$ \begin{aligned} t_C &= T - 273.15 \\ &= 216.55 - 273.15 \\ &= -56.60^\circ\text{C} \end{aligned} $$Convert Celsius → Fahrenheit
$$ \begin{aligned} t_F &= \frac{9}{5}t_C + 32 \\ &= \frac{9}{5}(-56.60) + 32 \\ &= -101.88 + 32 \\ &= -69.88^\circ\text{F} \end{aligned} $$Thus, the triple point of carbon dioxide is:
This diagram illustrates how the same physical temperature corresponds to different numerical values on the Kelvin, Celsius, and Fahrenheit scales.
Q2 Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between \(T_A\) and \(T_B\)?
An absolute temperature scale is a thermodynamic scale whose zero corresponds to absolute zero. All absolute scales are linearly proportional because they measure temperature from the same physical reference.
If two absolute scales are calibrated using the same fixed point (such as the triple point of water), their readings are related by:
$$ \frac{T_A}{(T_A)_{tp}} = \frac{T_B}{(T_B)_{tp}} $$where \( (T_A)_{tp} \) and \( (T_B)_{tp} \) represent the triple-point values on scales A and B.
Since both correspond to the same physical temperature, they can be equated through proportionality.
Thus, the temperature reading on scale A is \(\frac{4}{7}\) of the corresponding reading on scale B.
The diagram shows that although the numerical values differ (200 A and 350 B), both represent the same thermodynamic temperature (triple point of water), leading to a proportional relation between the two scales.
Q3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: \( R = R_0[1 + \alpha (T - T_0)] \). The resistance is \(101.6\,\Omega\) at the triple point of water \(273.16\,K\), and \(165.5\,\Omega\) at the normal melting point of lead \(600.5\,K\). What is the temperature when the resistance is \(123.4\,\Omega\)?
In a resistance thermometer, the electrical resistance of a metal changes approximately linearly with temperature over a moderate range. The relation is
$$ R = R_0 \left[1 + \alpha (T - T_0)\right] $$where
Using two known calibration points, we can first determine \(\alpha\), and then compute the unknown temperature.
The resistance thermometer shows an approximately linear increase of resistance with temperature. Using two calibration points, the temperature corresponding to any intermediate resistance can be determined.
Q4
Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why?
What is wrong in taking the melting point of ice and the boiling point of water as
standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale which were assigned
0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the
triple-point of water (273.16 K). What is the other fixed point on the Kelvin scale?
(c) The absolute temperature \(T\) (Kelvin scale) is related to the Celsius temperature \(t_c\) by
\(t_c = T - 273.15\). Why is the constant \(273.15\) used instead of \(273.16\)?
(d) What is the temperature of the triple-point of water on an absolute scale whose
unit interval size is equal to that of the Fahrenheit scale?
Thermometry requires fixed reference points that are reproducible and independent of environmental variations. Modern thermodynamic temperature scales therefore use the triple point of water and absolute zero as reference points because these temperatures are fundamental physical constants.
The triple point of water is the unique condition where ice, liquid water, and water vapour coexist in equilibrium. This occurs at a definite temperature and pressure.
This temperature is highly reproducible in laboratories.
In contrast:
Hence these points are not sufficiently stable for precise thermometry.
The central point represents the equilibrium of all three phases.
The Kelvin scale has two fundamental reference points:
Absolute zero corresponds to the lowest possible temperature where the thermal motion of particles becomes minimal.
$$ \boxed{\text{Other fixed point = Absolute Zero (0 K)}} $$The relation between Celsius and Kelvin scales is
$$ t_c = T - 273.15 $$This occurs because:
Thus the difference arises because the Celsius scale is referenced to the melting point of ice, not the triple point.
If an absolute scale uses the same unit size as Fahrenheit, its zero must still correspond to absolute zero.
Conversion between Kelvin and Fahrenheit intervals:
$$ 1\,K = \frac{9}{5} \,^\circ F = 1.8^\circ F $$Triple point of water:
$$ T = 273.16 K $$Therefore on this absolute Fahrenheit scale:
$$ \begin{aligned} T &= 273.16 \times 1.8 \\ T &\approx 491.7 \end{aligned} $$Q5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
| Temperature | Pressure (Thermometer A) | Pressure (Thermometer B) |
|---|---|---|
| Triple point of water | \(1.250 \times 10^5\ \text{Pa}\) | \(0.200 \times 10^5\ \text{Pa}\) |
| Normal melting point of sulphur | \(1.797 \times 10^5\ \text{Pa}\) | \(0.287 \times 10^5\ \text{Pa}\) |
(a) What is the absolute temperature of the normal melting point of sulphur as read by thermometers
A and B?
(b) Why do the readings differ slightly? What procedure reduces this discrepancy?
In a constant-volume gas thermometer, the pressure of a fixed mass of gas is directly proportional to its absolute temperature:
$$ T \propto P $$Thus, using the triple point of water as a reference:
$$ T = T_{tp}\frac{P}{P_{tp}} $$where
Given:
Given:
The difference occurs because real gases deviate slightly from ideal-gas behaviour. Oxygen and hydrogen exhibit different deviations from the ideal gas law at finite pressures, leading to slightly different temperature readings.
However, as pressure approaches zero, all gases behave more ideally.
The experiment should be repeated at progressively lower gas pressures, and the measured temperatures should be extrapolated to zero pressure.
At zero pressure, all gases approach ideal behaviour and yield the same thermodynamic temperature.
Different gases give slightly different slopes at finite pressure, but extrapolating to zero pressure yields the same thermodynamic temperature.
Q6 A steel tape 1 m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = \(1.20 \times 10^{-5}\ K^{-1}\).
When temperature changes, the length of a solid changes according to the law of linear expansion:
$$ L = L_0 (1 + \alpha \Delta T) $$where
If the measuring tape expands, the distance between its markings increases. Hence the scale reads a value slightly smaller than the true length.
Because the tape expands, the apparent length is related to the true length by
$$ L_{\text{apparent}} = \frac{L_{\text{true}}}{1+\alpha \Delta T} $$ Substituting values: $$ 63.0 = \frac{L}{1 + (1.20\times10^{-5})(18)} $$ $$ 63.0 = \frac{L}{1.000216} $$ Therefore, $$ L = 63.0 \times 1.000216 $$ $$ L \approx 63.014\ \text{cm} $$Thus, the actual length of the rod at \(45^\circ C\) is
$$ \boxed{63.014\ \text{cm}} $$Now the rod itself contracts when temperature decreases from \(45^\circ C\) to \(27^\circ C\).
Using $$ L = L_0 (1 + \alpha \Delta T) $$ Rearranging: $$ L_0 = \frac{L}{1 + \alpha \Delta T} $$ Substitute: $$ L_0 = \frac{63.014}{1.000216} $$ $$ L_0 \approx 63.000\ \text{cm} $$Hence, the length of the rod at \(27^\circ C\) is approximately
$$ \boxed{63.0\ \text{cm}} $$When the tape expands at higher temperature, the spacing between its markings increases. Hence the measured value becomes slightly smaller than the true length.
Q7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using dry ice. At what temperature of the shaft does the wheel slip on the shaft? Coefficient of linear expansion of steel: \( \alpha = 1.20 \times 10^{-5}\ K^{-1} \)
When temperature changes, the dimensions of a solid change according to the law of linear expansion:
$$ L = L_0 (1 + \alpha \Delta T) $$If the temperature decreases, the body undergoes thermal contraction. In mechanical engineering this method is called shrink fitting, where a shaft is cooled so that it contracts and fits into a hole.
For the wheel to slip onto the shaft, the shaft must contract until its diameter equals the hole diameter.
The shaft must be cooled to approximately
$$ \boxed{-69^\circ C} $$so that its diameter contracts sufficiently for the wheel to slip onto it.
Cooling the shaft reduces its diameter due to thermal contraction, allowing it to fit into the slightly smaller hole of the wheel.
Q8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper: \( \alpha = 1.70 \times 10^{-5}\ K^{-1} \)
When a metal sheet expands due to heating, every linear dimension increases in the same proportion. A useful principle in thermal expansion is:
A hole in a metal plate expands exactly as if the hole were filled with the same material.
Hence the diameter of the hole increases according to the law of linear expansion.
$$ L = L_0 (1 + \alpha \Delta T) $$ whereIncrease in diameter of the hole:
$$ \boxed{\Delta L = 1.44 \times 10^{-2}\ \text{cm}} $$Thus, heating the copper sheet to \(227^\circ C\) increases the diameter of the hole by approximately \(0.0144\ \text{cm}\).
When the sheet expands due to heating, the surrounding material moves outward, causing the hole to expand as well.
Q9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass: \( \alpha = 2.0 \times 10^{-5}\ K^{-1} \) Young’s modulus of brass: \( Y = 0.91 \times 10^{11}\ \text{Pa} \)
If a body is free, it contracts on cooling according to
$$ \frac{\Delta L}{L} = \alpha \Delta T $$However, when a wire is fixed between rigid supports, it cannot contract. This restriction produces thermal stress in the wire.
Using Young’s modulus:
$$ Y = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon} $$ Thus, $$ \sigma = Y\varepsilon $$ where the thermal strain is $$ \varepsilon = \alpha \Delta T $$The tension developed in the wire is approximately
$$ \boxed{F \approx 3.8 \times 10^{2}\ \text{N}} $$Since the wire cannot contract due to rigid supports, cooling produces tensile stress which results in a tension force in the wire.
Q10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a thermal stress developed at the junction? (Coefficient of linear expansion: brass \(= 2.0\times10^{-5}\ K^{-1}\), steel \(= 1.2\times10^{-5}\ K^{-1}\))
The change in length of a solid due to temperature change is given by the linear expansion relation:
$$ \Delta L = L_0 \alpha \Delta T $$where
If two rods are joined end-to-end but the ends are free, each rod expands independently according to its own coefficient of expansion. Hence, no thermal stress develops at the junction.
Since the ends of the combined rod are free to expand, both rods expand naturally according to their own coefficients of expansion.
Therefore,
$$ \boxed{\text{No thermal stress develops at the junction}} $$Because the ends are free, each rod expands independently and the junction does not experience thermal stress.
Q11 The coefficient of volume expansion of glycerine is \(49 \times 10^{-5}\ K^{-1}\). What is the fractional change in its density for a 30 °C rise in temperature?
When a liquid is heated, its volume increases according to the relation
$$ \frac{\Delta V}{V} = \beta \Delta T $$where
Density is defined as
$$ \rho = \frac{m}{V} $$Since the mass remains constant during heating, an increase in volume leads to a decrease in density. Thus,
$$ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} $$The fractional change in density is
$$ \boxed{\frac{\Delta \rho}{\rho} = -0.0147} $$Thus the density of glycerine decreases by approximately 1.47 % for a temperature rise of \(30^\circ C\).
Heating causes the liquid to expand (increase in volume), which reduces its density since the mass remains unchanged.
Q12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = \(0.91\ \text{J g}^{-1}\text{K}^{-1}\).
Mechanical work done by a machine can be converted into heat energy. The heat supplied to a body is related to the rise in temperature by
$$ Q = mc\Delta T $$where
If a machine of power \(P\) operates for time \(t\), the total energy supplied is
$$ E = Pt $$The rise in temperature of the aluminium block is
$$ \boxed{\Delta T \approx 103\ \text{K} \;(\text{or } 103^\circ C)} $$During drilling, mechanical work is converted into heat. A part of this heat raises the temperature of the aluminium block.
Q13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = \(0.39\ \text{J g}^{-1}\text{K}^{-1}\); Heat of fusion of ice = \(335\ \text{J g}^{-1}\))
In calorimetry, the heat lost by a hot body equals the heat gained by a colder body, provided there is no heat loss to the surroundings.
When the hot copper block is placed on ice at \(0^\circ C\), the copper cools down to \(0^\circ C\). The heat released during cooling is used to melt the ice.
The heat lost by copper: $$ Q = mc\Delta T $$ The heat required to melt ice: $$ Q = mL $$The maximum mass of ice that can melt is
$$ \boxed{1.45\ \text{kg}} $$The hot copper block releases heat while cooling to \(0^\circ C\). This heat is used to melt the ice without raising its temperature.
Q14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (water equivalent 0.025 kg) containing 150 cm³ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value?
In calorimetry experiments, the heat lost by the hot body equals the heat gained by the colder bodies, provided no heat is lost to the surroundings.
Thus, $$ \text{Heat lost by metal} = \text{Heat gained by water + calorimeter} $$ Using $$ Q = mc\Delta T $$If heat is lost to the surroundings, part of the heat released by the metal does not heat the water and calorimeter. Therefore the calculated heat gain becomes smaller.
Hence the calculated value of \(c\) becomes smaller than the true value.
Heat flows from the hot metal to the water and calorimeter until thermal equilibrium is reached.
Q15 Given below are observations on molar specific heats at room temperature of some common gases.
| Gas | Molar specific heat \(C_v\) (cal mol\(^{-1}\) K\(^{-1}\)) |
|---|---|
| Hydrogen | 4.87 |
| Nitrogen | 4.97 |
| Oxygen | 5.02 |
| Nitric oxide | 4.99 |
| Carbon monoxide | 5.01 |
| Chlorine | 6.17 |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, the molar specific heat of a monatomic gas is 2.92 cal mol\(^{-1}\) K\(^{-1}\). Explain this difference. What can you infer from the somewhat larger value for chlorine?
According to the equipartition theorem, each degree of freedom contributes \(\tfrac{1}{2}R\) per mole to the molar specific heat at constant volume.
Thus, if a molecule has \(f\) degrees of freedom,
$$ C_v = \frac{f}{2}R $$A monatomic gas atom can move only in three perpendicular directions (translational motion).
Number of degrees of freedom: $$ f = 3 $$ Therefore, $$ C_v = \frac{3}{2}R $$ Using \(R = 1.987\ \text{cal mol}^{-1}\text{K}^{-1}\): $$ C_v = \frac{3}{2} \times 1.987 $$ $$ C_v \approx 2.98\ \text{cal mol}^{-1}\text{K}^{-1} $$ which is close to the observed value \(2.92\).Diatomic molecules have additional rotational motion.
Active degrees of freedom at room temperature:This agrees well with the observed values for H\(_2\), N\(_2\), O\(_2\), NO, and CO.
The molar specific heat of chlorine is slightly larger (\(6.17\ \text{cal mol}^{-1}\text{K}^{-1}\)).
This suggests that at room temperature some vibrational degrees of freedom are also partially excited in chlorine molecules.
Vibrational motion adds additional energy storage, increasing the molar specific heat.
Q16 A child running a temperature of 101°F is given an antipyrin (a medicine that lowers fever) which increases the rate of evaporation of sweat from the body. If the fever falls to 98°F in 20 minutes, what is the average rate of extra evaporation caused by the drug? Assume evaporation is the only way heat is lost. Mass of child = 30 kg. Specific heat of human body ≈ specific heat of water. Latent heat of evaporation of water = 580 cal g\(^{-1}\).
Human bodies cool mainly through evaporation of sweat. When sweat evaporates, it absorbs latent heat from the body, thereby lowering body temperature.
Heat required for evaporation: $$ Q = mL $$ Heat lost due to temperature fall: $$ Q = mc\Delta T $$Average rate of extra evaporation caused by the drug:
$$ \boxed{ \approx 4.3\ \text{g min}^{-1} } $$Evaporation of sweat removes heat from the body because latent heat is required to convert liquid sweat into vapour.
Q17 A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. Outside temperature = \(45^\circ C\) Thermal conductivity of thermacole \(k = 0.01\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}\) Heat of fusion of ice \(L = 335 \times 10^3\ \text{J kg}^{-1}\)
Heat flows through the walls of the icebox by thermal conduction. The rate of heat transfer through a slab is given by
$$ \frac{Q}{t} = \frac{kA\Delta T}{x} $$ whereHeat from the hot surroundings slowly conducts through the thermacole walls and melts part of the ice inside the box.
Q18 A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass \(k = 109\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}\) Heat of vaporisation of water \(L = 2256 \times 10^{3}\ \text{J kg}^{-1}\)
Q19
Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer calibrated for an ideal black body gives too low a value
for the temperature of a red-hot iron piece in the open, but gives the correct
temperature when the same piece is in a furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming
a building than those based on circulation of hot water
Thermal radiation and heat transfer depend on properties such as emissivity, absorptivity, reflectivity, thermal conductivity, and latent heat. Kirchhoff’s law of radiation states that good absorbers are also good emitters at a given temperature.
If a body reflects most of the radiation incident on it, it absorbs very little energy. According to Kirchhoff’s law of thermal radiation,
$$ \text{Good absorber} \Rightarrow \text{Good emitter} $$Therefore, a body that absorbs little radiation (high reflectivity) also emits little radiation, making it a poor emitter.
Although both objects may be at the same temperature, brass is a good conductor of heat while wood is a poor conductor.
An optical pyrometer determines temperature by comparing radiation intensity with that of an ideal black body.
Inside a furnace, multiple reflections make the object behave almost like a black body, so the pyrometer reads the correct temperature.
The atmosphere absorbs part of the infrared radiation emitted by the earth and re-radiates it back to the surface.
This greenhouse effect reduces heat loss to space. Without the atmosphere:
Hence the earth would become extremely cold and unsuitable for life.
Steam releases a large amount of heat when it condenses into water. This heat is called latent heat of condensation.
Q20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
According to Newton’s law of cooling, the rate of loss of heat of a body is proportional to the difference between its temperature and the surrounding temperature.
Mathematically, $$ T - T_s = (T_0 - T_s)e^{-kt} $$ whereThe temperature difference between the body and surroundings decreases exponentially with time according to Newton’s law of cooling.
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