Heat and Thermodynamics

61. Molar heat capacity at constant pressure of a diatomic gas is 3·5 R, w

Molar heat capacity at constant pressure of a diatomic gas is 3·5 R, where R is universal gas constant. The gas is heated by providing energy of 1400 J and is allowed to expand isobarically. Which one among the following represents the work done by the gas in this process ?

[amp_mcq option1=”400 J” option2=”4900 J” option3=”560 J” option4=”280 J” correct=”option1″]

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The correct option is A.
For an ideal gas, the molar heat capacities at constant pressure (C_p) and constant volume (C_v) are related by Mayer’s relation: C_p – C_v = R. The work done by a gas during an isobaric (constant pressure) process is given by W = PΔV = nRΔT. The heat added during an isobaric process is Q = nC_pΔT. By the First Law of Thermodynamics, Q = ΔU + W, where ΔU = nC_vΔT.
Given molar heat capacity at constant pressure C_p = 3.5 R = (7/2)R.
Using Mayer’s relation, C_v = C_p – R = (7/2)R – R = (5/2)R.
The gas is heated isobarically, and energy Q = 1400 J is provided.
For an isobaric process, Q = nC_pΔT, where n is the number of moles and ΔT is the temperature change.
So, 1400 J = n * (7/2)R * ΔT.
By the First Law of Thermodynamics, Q = ΔU + W.
Work done by the gas W = Q – ΔU.
Change in internal energy for an ideal gas ΔU = nC_vΔT.
W = Q – nC_vΔT.
From the isobaric heat equation, nΔT = Q / C_p = 1400 / (7/2)R = 1400 * (2 / 7R) = 2800 / 7R = 400 / R.
Now substitute this into the work equation:
W = Q – nC_vΔT = 1400 – (nΔT) * C_v = 1400 – (400/R) * (5/2)R
W = 1400 – 400 * (5/2) = 1400 – 200 * 5 = 1400 – 1000 = 400 J.

Alternatively, W = nRΔT.
From Q = nC_pΔT, we have nΔT = Q/C_p.
W = R * (Q/C_p) = Q * (R/C_p).
Given C_p = (7/2)R, so R/C_p = R / ((7/2)R) = 2/7.
W = 1400 J * (2/7) = 200 * 2 = 400 J.

62. An ice bullet of temperature – 10°C is fired upon an object. The bulle

An ice bullet of temperature – 10°C is fired upon an object. The bullet spends 20% of its kinetic energy in penetrating into the object and remaining kinetic energy is spent into melting the bullet at 0°C. The latent heat of fusion of ice is 3·4 × 10⁵ J kg⁻¹ and its specific heat capacity is 2000 J kg⁻¹ K⁻¹. Which one among the following is the correct initial speed of the bullet ?

[amp_mcq option1=”300√10 m s⁻¹” option2=”3600 m s⁻¹” option3=”300√5 m s⁻¹” option4=”900 m s⁻¹” correct=”option1″]

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The correct option is A.
The initial kinetic energy of the ice bullet is converted into different forms of energy upon impact: a portion for penetration and the rest for heating the ice to its melting point (0°C) and then melting it at 0°C.
Let m be the mass of the ice bullet and v be its initial speed.
Initial kinetic energy (KE) = (1/2)mv².
20% of KE is spent on penetration = 0.2 * (1/2)mv² = 0.1mv².
Remaining KE is spent on heating the ice from -10°C to 0°C and melting it at 0°C.
Remaining KE = KE – 0.1mv² = 0.9mv².
Energy required to heat mass m of ice from -10°C to 0°C:
Q_heat = m * c_ice * ΔT = m * (2000 J kg⁻¹ K⁻¹) * (0 – (-10)) K = m * 2000 * 10 = 20000m J.
Energy required to melt mass m of ice at 0°C:
Q_melt = m * L = m * (3.4 × 10⁵ J kg⁻¹) = 340000m J.
Total energy required for heating and melting = Q_heat + Q_melt = 20000m + 340000m = 360000m J.
This energy comes from the remaining KE:
0.9mv² = 360000m
0.9v² = 360000
v² = 360000 / 0.9 = 3600000 / 9 = 400000.
v = √400000 = √((4 × 10⁵) * 10/10) = √(40 × 10⁴) = 100 * √40 = 100 * √(4 * 10) = 100 * 2√10 = 200√10 m/s.

Wait, let’s re-read the problem carefully: “remaining kinetic energy is spent into melting the bullet at 0°C”. This phrasing usually means only the phase change energy is considered from the remaining KE. If this interpretation is correct, the 80% KE is used *only* for melting at 0°C, implying the energy for heating from -10°C to 0°C is either negligible or sourced otherwise (e.g., from the 20% penetration energy, which is unlikely). Let’s retry with this interpretation:

Remaining KE (80% of total KE) = 0.8 * (1/2)mv² = 0.4mv².
This is spent on melting the bullet at 0°C:
0.4mv² = Q_melt = m * L = m * 3.4 × 10⁵
0.4v² = 3.4 × 10⁵
v² = (3.4 × 10⁵) / 0.4 = (3.4 × 10⁵) / (4/10) = (3.4 × 10⁵) * (10/4) = (34/4) × 10⁵ = 8.5 × 10⁵
v = √(8.5 × 10⁵) = √(85 × 10⁴) = 100 * √85.
√85 is approximately 9.22. v ≈ 922 m/s. This doesn’t match any option well.

Let’s go back to the first interpretation, where 80% KE covers both heating and melting.
0.8 * (1/2)mv² = m * c_ice * ΔT + m * L
0.4v² = m(c_ice * ΔT + L)
0.4v² = 2000 * 10 + 3.4 × 10⁵ = 20000 + 340000 = 360000
v² = 360000 / 0.4 = 900000
v = √900000 = √(90 × 10⁴) = 100 * √90 = 100 * √(9 × 10) = 100 * 3√10 = 300√10 m/s.
300√10 ≈ 300 * 3.162 = 948.6 m/s. This matches option A exactly.

The phrase “melting the bullet at 0°C” in the problem statement, combined with providing the specific heat capacity of ice and the initial temperature of -10°C, strongly suggests that the energy required to raise the temperature to 0°C must also be accounted for by the available KE. Therefore, the 80% KE is used for the entire process from -10°C to liquid water at 0°C (heating ice + melting ice).

63. A cup of tea is placed at a table, and the room temperature is 35°C. T

A cup of tea is placed at a table, and the room temperature is 35°C. The tea cools from 90°C to 70°C in 5 minutes. Which one among the following is the correct time taken for the tea to cool from 70°C to 50°C ?

[amp_mcq option1=”5 minutes” option2=”7 minutes” option3=”9 minutes” option4=”11 minutes” correct=”option3″]

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The correct option is C.
According to Newton’s Law of Cooling, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. That is, dθ/dt ∝ (θ – θ₀), where θ is the object’s temperature and θ₀ is the surrounding temperature. For small temperature ranges, the average rate of cooling in an interval can be approximated as proportional to the average temperature difference in that interval.
Let the surrounding temperature be θ₀ = 35°C.
In the first interval, the tea cools from 90°C to 70°C in t₁ = 5 minutes.
The temperature change is Δθ₁ = 90 – 70 = 20°C.
The average temperature in this interval is θ_avg₁ = (90 + 70) / 2 = 80°C.
The average temperature difference is ΔT₁ = θ_avg₁ – θ₀ = 80 – 35 = 45°C.
The average rate of cooling in this interval is approximately Rate₁ ≈ Δθ₁ / t₁ = 20°C / 5 min = 4°C/min.
By Newton’s Law of Cooling, Rate₁ ∝ ΔT₁, so 4 ∝ 45.

In the second interval, the tea cools from 70°C to 50°C. Let the time taken be t₂.
The temperature change is Δθ₂ = 70 – 50 = 20°C.
The average temperature in this interval is θ_avg₂ = (70 + 50) / 2 = 60°C.
The average temperature difference is ΔT₂ = θ_avg₂ – θ₀ = 60 – 35 = 25°C.
The average rate of cooling in this interval is approximately Rate₂ ≈ Δθ₂ / t₂ = 20°C / t₂.
By Newton’s Law of Cooling, Rate₂ ∝ ΔT₂, so (20/t₂) ∝ 25.

Taking the ratio of the rates:
Rate₁ / Rate₂ = ΔT₁ / ΔT₂
(4) / (20/t₂) = 45 / 25
4t₂ / 20 = 9 / 5
t₂ / 5 = 9 / 5
t₂ = 9 minutes.
The time taken for the tea to cool from 70°C to 50°C is 9 minutes.

64. When the temperature of a gas increases, the average speed of its

When the temperature of a gas increases, the average speed of its molecules

[amp_mcq option1=”remains the same” option2=”decreases” option3=”increases” option4=”either increases or decreases depending on the gas” correct=”option3″]

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The correct answer is C) increases.
According to the kinetic theory of gases, the absolute temperature of an ideal gas is directly proportional to the average kinetic energy of its molecules. Kinetic energy is given by the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass of a molecule and $v$ is its speed. As temperature increases, the average kinetic energy of the molecules increases. Since the mass of the molecules remains constant, an increase in average kinetic energy implies an increase in the average value of $v^2$. The average speed of the molecules is related to the average of the square of the speed (root-mean-square speed), and both increase with increasing temperature.
The root-mean-square speed ($v_{rms}$) of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the absolute temperature, and $m$ is the mass of a molecule. This formula clearly shows that the average speed of gas molecules increases with the square root of the absolute temperature. Therefore, as the temperature of a gas increases, the average speed of its molecules also increases.

65. A sealed container contains one mole of helium and three moles of nitr

A sealed container contains one mole of helium and three moles of nitrogen gas at 25 °C. If the total pressure is 80000 Pa, what will be the partial pressure of nitrogen gas in the container?

[amp_mcq option1=”20000 Pa” option2=”40000 Pa” option3=”60000 Pa” option4=”80000 Pa” correct=”option3″]

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The correct answer is 60000 Pa.
According to Dalton’s Law of Partial Pressures, the partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure of the mixture.
Total moles of gas = Moles of Helium + Moles of Nitrogen = 1 mole + 3 moles = 4 moles. The mole fraction of nitrogen is (Moles of Nitrogen) / (Total moles) = 3 / 4. The total pressure is 80000 Pa. Partial pressure of nitrogen = (Mole fraction of nitrogen) * (Total pressure) = (3/4) * 80000 Pa = 3 * 20000 Pa = 60000 Pa.

66. The speed of particles in a matter usually increases with a rise in te

The speed of particles in a matter usually increases with a rise in temperature. Which one of the following phenomena is responsible for it?

[amp_mcq option1=”An increase in kinetic energy” option2=”A decrease in kinetic energy” option3=”An increase in decomposition” option4=”A decrease in reaction rate” correct=”option1″]

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The phenomenon responsible for the speed of particles in matter usually increasing with a rise in temperature is an increase in kinetic energy.
– Temperature is a direct measure of the average kinetic energy of the particles (atoms, molecules, ions) within a substance.
– As temperature increases, the particles absorb thermal energy, which is converted into kinetic energy.
– Increased kinetic energy means the particles move faster and vibrate more vigorously.
– This increase in particle speed and kinetic energy explains phenomena like diffusion happening faster at higher temperatures, the expansion of matter when heated, and the transition between states of matter (solid to liquid, liquid to gas) as particles gain enough energy to overcome intermolecular forces.
– Options C and D relate to chemical reaction rates, which are influenced by the kinetic energy and collisions of particles, but the fundamental reason for increased particle speed with temperature is the increased kinetic energy itself.

67. A copper rod of length 1 m is initially at room temperature. If the te

A copper rod of length 1 m is initially at room temperature. If the temperature of the rod is increased by 100 °C, what will be the change in length of the rod? (Given linear thermal expansion coefficient of Cu is 17 x 10⁻⁶ / °C)

[amp_mcq option1=”0.17 mm” option2=”1.7 mm” option3=”0.17 cm” option4=”1.7 cm” correct=”option2″]

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The change in length of the copper rod will be 1.7 mm.
The change in length due to thermal expansion is given by the formula ΔL = L₀ * α * ΔT.
Given: L₀ = 1 m = 1000 mm, α = 17 x 10⁻⁶ / °C, ΔT = 100 °C.
ΔL = 1000 mm * (17 x 10⁻⁶ / °C) * 100 °C
ΔL = 1000 * 17 * 10⁻⁶ * 100 mm
ΔL = 1700000 x 10⁻⁶ mm
ΔL = 1.7 mm.

68. At atmospheric pressure, the density of water is maximum at a temperat

At atmospheric pressure, the density of water is maximum at a temperature of:

[amp_mcq option1=”0 °C” option2=”4 °C” option3=”-4 °C” option4=”1 °C” correct=”option2″]

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At atmospheric pressure, the density of water is maximum at a temperature of 4 °C. This is due to the anomalous expansion of water. As water is heated from 0 °C to 4 °C, it contracts (density increases), unlike most substances which expand upon heating. Above 4 °C, water expands normally upon heating (density decreases). Conversely, when water is cooled from higher temperatures, its density increases until it reaches a maximum at 4 °C, and then decreases as it is cooled from 4 °C to 0 °C (it expands upon freezing at 0 °C).
Water exhibits anomalous expansion, reaching its maximum density at 4 °C.
This peculiar property of water is crucial for aquatic life in cold climates, as the densest water (at 4 °C) settles at the bottom of lakes and ponds, preventing them from freezing solid from top to bottom and allowing aquatic organisms to survive through the winter.

69. In a thermos flask, the walls are made of shiny glass. Which one of th

In a thermos flask, the walls are made of shiny glass. Which one of the following heat loss process is minimised by using shiny walls?

[amp_mcq option1=”Convection” option2=”Conduction” option3=”Radiation” option4=”Both radiation and convection” correct=”option3″]

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In a thermos flask, the walls are made of shiny glass to minimise heat loss by radiation. Shiny surfaces are poor emitters and poor absorbers of thermal radiation. By making the inner and outer surfaces shiny, heat transfer by radiation is significantly reduced. Heat from the hot liquid inside is not easily radiated outwards, and external heat is not easily absorbed and radiated inwards.
Shiny surfaces are poor emitters and absorbers of thermal radiation, effectively reducing heat transfer by radiation.
A thermos flask is designed to minimise all three modes of heat transfer: conduction (by using a vacuum between the walls and a stopper), convection (by using a vacuum and a narrow neck), and radiation (by using shiny surfaces and a vacuum). The shiny walls specifically target the reduction of heat transfer through radiation.

70. Which one of the following heat transfers is an example of convection

Which one of the following heat transfers is an example of convection ?

[amp_mcq option1=”Heating of food in a microwave oven” option2=”Boiling water in a pot on a gas stove” option3=”Feeling the warmth in sun” option4=”Heating a brass rod at one end and observing the temperature rise at the other end” correct=”option2″]

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Boiling water in a pot on a gas stove is an example of heat transfer primarily by convection within the water.
Convection is the transfer of heat through the movement of a fluid (liquid or gas). In boiling water, heat from the stove is conducted through the pot to the water at the bottom. This heated water becomes less dense and rises, while cooler, denser water from the top sinks, creating circulating currents (convection currents) that distribute heat throughout the liquid.
Heat transfer in a microwave oven (A) is primarily through dielectric heating caused by microwave radiation exciting water molecules. Feeling warmth from the sun (C) is heat transfer by radiation through electromagnetic waves. Heating a brass rod (D) involves heat transfer by conduction through the solid material from the hotter end to the cooler end via molecular vibrations and free electron movement.
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