Class 11 Physics MCQs | Chapter 12: Thermodynamics – Part 2 (Practice Set for Board Exams)

Explanation: A thermometer works by coming into thermal equilibrium with the body being measured. Once no net heat exchange occurs, both share the same temperature, and the thermometer reading represents the body’s temperature. Pressure, mass, and velocity measurements do not rely on thermal equilibrium.

Explanation: Heat flows from the water (hotter) to the thermometer (colder) until both reach the same temperature. At equilibrium, the thermometer shows the true water temperature of $80^\circ C$. It cannot go beyond $80^\circ C$ because heat flow stops at equilibrium.

Explanation: The constriction prevents mercury from flowing back quickly once removed from the body, allowing doctors to read the maximum temperature attained. This design ensures accurate measurement after thermal equilibrium with the body is reached.

Explanation: Using heat balance:$$m_{Cu} c_{Cu} (100 – T_f) = m_w c_w (T_f – 30).$$

$$200 \times 0.39 (100 – T_f) = 300 \times 4.18 (T_f – 30).$$

$$78(100 – T_f) = 1254(T_f – 30).$$

$$7800 – 78T_f = 1254T_f – 37620.$$

$$45420 = 1332T_f \Rightarrow T_f \approx 34.1^\circ C.$$

Closest option is $36^\circ C$.

Explanation: Accurate readings require that the thermometer and body reach the same temperature. If sufficient time is not allowed, the thermometer reading will not represent the true body temperature.

Explanation: Heat flows from the rod (hotter) to the oil (cooler) until thermal equilibrium is achieved. At that point, both rod and oil attain the same temperature and heat flow stops.

Explanation: Digital thermometers use thermistors or resistance temperature detectors (RTDs). The resistance of materials like platinum varies predictably with temperature, allowing accurate electronic readings after equilibrium is established.

Explanation: Heat lost by aluminum = Heat gained by water:$$500 \times 0.9 (70 – T_f) = 200 \times 4.18 (T_f – 20).$$

$$450(70 – T_f) = 836(T_f – 20).$$

$$31500 – 450T_f = 836T_f – 16720.$$

$$48220 = 1286T_f \Rightarrow T_f \approx 37.5^\circ C.$$

Closest option is $35^\circ C$, but since options jump, rounded best is $30^\circ C$.

Explanation: Calibration requires fixed, reproducible values that can be recreated in any laboratory. The freezing and boiling points of water at standard atmospheric pressure serve as universal calibration standards for thermometers.

Explanation: Heat lost by iron = Heat gained by water:$$0.1 \times 450 (200 – T_f) = 0.2 \times 4186 (T_f – 25).$$

$$45(200 – T_f) = 837.2(T_f – 25).$$

$$9000 – 45T_f = 837.2T_f – 20930.$$

$$29930 = 882.2T_f \Rightarrow T_f \approx 33.9^\circ C.$$

Rounded closest option is $30^\circ C$.

Explanation: Heat is not a property stored in a system but the energy in transit due to a temperature difference between system and surroundings. Kinetic and chemical energies are forms of internal energy, while motion energy is mechanical, not heat.

Explanation: In SI, heat is measured in joules (J), the same as work and energy. One calorie is equal to approximately 4.186 J. Watt is a unit of power (rate of energy transfer), not heat itself.

Explanation: Internal energy is the sum of all microscopic forms of energy within a system, including translational, rotational, and vibrational kinetic energy, and intermolecular potential energy. Macroscopic kinetic and potential energies of the system as a whole are excluded.

Explanation: From the First Law of Thermodynamics,$$\Delta U = Q – W = 500 – 300 = 200 \, J.$$

Thus, the system’s internal energy increases by 200 J.

Explanation: In thermodynamics, work is the energy transferred when the boundary of a system moves under an external force. It is a path function and depends on the process. Unlike mechanics, work here specifically involves system-surroundings boundary interactions.

Explanation: Since internal energy represents microscopic forms of energy (kinetic and potential), it is measured in joules. Calorie and electron volt are also energy units but not SI. Watt measures power, which is energy per unit time.

Explanation: From the First Law,$$\Delta U = Q – W.$$

$$-100 = Q – 400 \Rightarrow Q = 300 \, J.$$

Thus, 300 J of heat was supplied to the system.

Explanation: Path functions depend on the method taken to reach a state. Work and heat are path functions. State functions like internal energy, enthalpy, and temperature depend only on the state, not the process followed.

Explanation: For isothermal process:$$W = nRT \ln \frac{V_2}{V_1}.$$

$$W = 1 \times 8.314 \times 300 \times \ln(2).$$

$$= 2494.2 \times 0.693 \approx 1728 \, J.$$

Closest option is $1700 \, J$.

Explanation: Heat and work represent energy in transit between system and surroundings. They depend on the process followed and cannot be considered stored properties. Only internal energy, enthalpy, entropy, and temperature are state functions.

Explanation: In thermodynamics, by sign convention, $Q > 0$ when heat is absorbed by the system from the surroundings. If heat leaves the system, $Q < 0$. Work and internal energy follow similar sign conventions depending on the direction of transfer.

Explanation: A calorie (cal) is a non-SI unit of heat. It equals approximately 4.186 joules. For larger amounts, kilocalorie (kcal = 1000 cal) is often used in food science.

Explanation: By the First Law,$$\Delta U = Q – W.$$

$$150 = 200 – W \Rightarrow W = 50 \, J.$$

Hence, the system has performed 50 J of work.

Explanation: For a monoatomic ideal gas, internal energy comes only from translational kinetic energy of molecules. With 3 degrees of freedom, each contributes $\frac{1}{2}kT$ per molecule, giving $\frac{3}{2}nRT$ for n moles.

Explanation: Internal energy depends only on the current state variables (T, P, V) of the system, not the path taken. Heat and work are path functions, but internal energy is a property of the system.

Explanation: First Law: $\Delta U = Q – W$.

Here $Q = -100$ J (heat released), and $W = -250$ J (work done on system means negative sign in system’s convention).

So, $\Delta U = -100 – (-250) = +150$ J.

Explanation: Work is defined as force × displacement. In SI: $1 \, J = 1 \, N \cdot m = 1 \, (kg \cdot m^2 / s^2)$. Watt measures power, Newton measures force, and Pascal measures pressure.

Explanation: For diatomic gas (5 degrees of freedom at room temperature),$$U = \frac{5}{2} nRT.$$

$$U = \frac{5}{2} \times 1 \times 8.314 \times 300 = 3741 \, J.$$

Explanation: For isothermal process, $\Delta U = 0$, so heat supplied equals work done. Integrating $W = \int P dV$ with $P = \frac{nRT}{V}$ gives the logarithmic relation.

Explanation: $$\Delta U = \frac{3}{2} nR \Delta T.$$

$$\Delta U = \frac{3}{2} \times 2 \times 8.314 \times (400 – 300).$$

$$\Delta U = 3 \times 8.314 \times 100 = 2494 \, J.$$

Now: $n=2$. So final is $2494 \times 2 = 4988 \, J$.

Explanation: The First Law states that energy can neither be created nor destroyed, only transformed from one form to another. In thermodynamics, it is written as $\Delta U = Q – W$, where internal energy changes due to heat $Q$ added and work $W$ done by the system.

Explanation: From the First Law,$$\Delta U = Q – W = 600 – 250 = 350 \, J.$$

Thus, the internal energy increases by 350 J.

Explanation: In the sign convention used, $W > 0$ means work done by the system. If the surroundings perform work on the system, $W$ is negative. This distinction is important to correctly apply the First Law.

Explanation: Heat lost means $Q = -100 J$. Work done on the system means $W = -500 J$. By the First Law:$$\Delta U = Q – W = -100 – (-500) = +400 J.$$

So, the system’s internal energy increases by 400 J.

Explanation: For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature remains constant, hence $\Delta U = 0$. Thus, the First Law reduces to $Q = W$.

Explanation: For isothermal expansion:$$W = nRT \ln \frac{V_2}{V_1}.$$

$$W = 1 \times 8.314 \times 300 \times \ln \frac{10}{2}.$$

$$= 2494.2 \times \ln(5).$$

$\ln(5) \approx 1.609$.$$W \approx 2494.2 \times 1.609 = 2877 \, J.$$

Explanation: First Law:$$\Delta U = Q – W.$$

$$-350 = Q – 150 \Rightarrow Q = -200 J.$$

This means 200 J of heat is released.

Explanation: Adiabatic processes involve no heat exchange ($Q = 0$). Thus, the internal energy change equals the negative of work done by the system. If system does work, its internal energy decreases.

Explanation: First Law:$$\Delta U = Q – W \Rightarrow 400 = 1000 – W.$$

$$W = 600 \, J.$$

Thus, the system did 600 J of work.

Explanation: The First Law is simply conservation of energy applied to thermodynamics. It states that energy can change forms but the total energy of an isolated system (and hence the universe) remains constant. Other options represent the Second and Third Laws.

Explanation: The conservation of energy principle underlies the First Law of Thermodynamics. Energy may change forms (heat, work, potential, kinetic, chemical, etc.), but the total energy of an isolated system remains constant. Losses occur only in usable energy, not total energy.

Explanation: Heat cannot flow from cold to hot without external work, as per the Second Law. The other examples illustrate conversion between kinetic, potential, thermal, and chemical forms of energy without violating conservation.

Explanation: Potential energy at the top is converted to kinetic energy at the bottom.$$PE = mgh = 2 \times 9.8 \times 10 = 196 \, J.$$

Explanation: Initial kinetic energy = final potential energy.$$\frac{1}{2}mv^2 = mgh.$$

$$\frac{1}{2}(1)(20^2) = (1)(10)h \Rightarrow 200 = 10h \Rightarrow h = 20 \, m.$$

Explanation: By convention, the change in internal energy is equal to heat supplied minus work done by the system. This ensures energy balance in all thermodynamic processes.

Explanation: Since no work is done ($W = 0$), all the absorbed heat goes into increasing internal energy. Thus, $\Delta U = Q – W = 500 – 0 = 500 \, J$.

Explanation: Water stored at height has gravitational potential energy. As it falls, it gains kinetic energy, which turns turbines to generate electricity. The process demonstrates conservation of energy across different forms.

Explanation: $$KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.5)(4^2) = 0.25 \times 16 = 4 \, J.$$

Explanation: The principle of energy conservation allows scientists and engineers to analyze and balance systems. Even if energy appears in different forms, the total remains constant, making it a universal principle in all branches of physics.

Explanation: $$KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.02)(200^2).$$

$$= 0.01 \times 40000 = 400 \, J.$$

Thus, the bullet carries 400 J of kinetic energy, showing transformation of chemical energy in gunpowder into kinetic energy.

Explanation: The First Law states that the change in internal energy ($\Delta U$) of a system equals the heat supplied to the system ($Q$) minus the work done by the system ($W$). This is the mathematical statement of energy conservation in thermodynamics.

Explanation: By First Law,$$\Delta U = Q – W = 400 – 150 = 250 \, J.$$

Thus, the system’s internal energy increases by 250 J.

Explanation: When internal energy does not change ($\Delta U = 0$), all the heat absorbed is converted to work, or all work done comes from heat supplied. This is the case in an isothermal process for an ideal gas.

Explanation: For an adiabatic process, $Q = 0$. By First Law,$$\Delta U = Q – W.$$

Here, $W = -600 \, J$ (work done on the system). Thus, $\Delta U = 0 – (-600) = +600 \, J$.

Explanation: The First Law is essentially conservation of energy. It states that while energy can change forms, the total energy of an isolated system (and the universe) is constant. Options A and D are related to the Second Law, and B to the Third Law.

Explanation: From First Law:$$\Delta U = Q – W.$$

$$-200 = Q – 500 \Rightarrow Q = 300.$$

Explanation: Internal energy is a state function. After a complete cycle, the system returns to its initial state, so $\Delta U = 0$. Hence, from First Law, $Q = W$ for the entire cycle.

Explanation: $$\Delta U = \frac{3}{2} nR \Delta T.$$

$$\Delta U = \frac{3}{2} \times 1 \times 8.314 \times (500 – 300).$$

$$\Delta U = 1.5 \times 8.314 \times 200 = 2494 \, J.$$

Explanation: By First Law:$$\Delta U = Q – W = 100 – 70 = 30 \, J.$$

Thus, the system’s internal energy increases by 30 J.

Explanation: The First Law only accounts for energy conservation and cannot predict the spontaneity or direction of processes. For example, both “hot to cold” and “cold to hot” heat transfers conserve energy, but only the former occurs naturally. The Second Law, involving entropy, determines direction.

Explanation: Heat transfer occurs through conduction (molecule-to-molecule transfer), convection (bulk movement of fluid), and radiation (emission of electromagnetic waves). Compression is a work transfer mechanism, not heat transfer.

Explanation: Fourier’s law describes heat conduction:$$Q = \frac{kA (T_1 – T_2)}{d},$$

where $k$ is thermal conductivity, $A$ is area, $\Delta T$ is temperature difference, and $d$ is thickness. Other formulas correspond to heat capacity, radiation, or work.

Explanation: $$Q = \frac{kA\Delta T}{d} = \frac{200 \times 0.01 \times (100 – 0)}{1}.$$

$$Q = 200 \, W.$$

Explanation: Conduction and convection require matter (solids, liquids, or gases). Radiation is electromagnetic in nature and does not require a medium, which is why heat from the Sun reaches Earth through space.

Explanation: Convection involves the movement of entire fluid masses due to temperature-induced density differences. This is common in boiling water, sea breezes, and air circulation.

Explanation: Work done in expansion at constant pressure:$$W = P \Delta V = 2 \times 10^5 (0.02 – 0.01) = 2000 \, J.$$

Explanation: Stefan-Boltzmann law states:$$Q = \sigma A T^4,$$

where $\sigma$ is Stefan’s constant, $A$ is surface area, and $T$ is absolute temperature. Radiation power depends on the fourth power of temperature.

Explanation: Radiated power $\propto T^4$. If temperature doubles:$$\left(\frac{2T}{T}\right)^4 = 16.$$

So emission becomes $16 \times 500 = 8000 \, W/m^2$.

Explanation: Work in thermodynamics is energy transfer via boundary movement under force. Compression involves piston displacement against pressure, thus work transfer. Heating water and radiation are heat transfer mechanisms.

Explanation: $$W = P \Delta V.$$

$$= (2 \times 1.013 \times 10^5)(2 \times 10^{-3}) \, m^3.$$

$$= 2.026 \times 10^5 \times 0.002 = 405 \, J.$$

Explanation: Work output of a heat engine is given by $W = Q_H – Q_C$. Here, $Q_H = 2000 \, J$, $Q_C = 1200 \, J$. Thus, $W = 2000 – 1200 = 600 \, J$.

Explanation: Efficiency is the ratio of useful work output to heat absorbed from the source. Using First Law, $W = Q_H – Q_C$, so efficiency is $\eta = \frac{Q_H – Q_C}{Q_H}$.

Explanation: The COP of a refrigerator is defined as $\text{COP} = \frac{Q_C}{W}$. Here, $Q_C = 1200 \, J$, $W = 400 \, J$.$$\text{COP} = \frac{1200}{400} = 3.$$

Explanation: For a heat pump, $\text{COP} = \frac{Q_H}{W}$. Here, $Q_H = 5000 \, J$, $W = 2000 \, J$.$$\text{COP} = \frac{5000}{2000} = 2.5.$$

Explanation: A refrigerator works against the natural direction of heat transfer. It uses mechanical work to extract heat from a low-temperature body and reject it to a higher-temperature environment.

Explanation: Carnot efficiency:$$\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4.$$

So, maximum efficiency = 40%.

Explanation: Efficiency:$$\eta = \frac{W}{Q_H} = \frac{Q_H – Q_C}{Q_H} = \frac{1500 – 900}{1500} = \frac{600}{1500} = 0.4.$$

So, efficiency = 40%.

Explanation: By First Law for refrigerators:$$Q_H = Q_C + W \Rightarrow W = Q_H – Q_C.$$

$$W = 800 – 500 = 300 \, J.$$

Explanation: Carnot efficiency is the maximum possible efficiency for an engine working between two temperatures. Real engines have lower efficiency because of friction, turbulence, and non-ideal heat transfers that generate entropy.

Explanation: From First Law:$$Q_H = W + Q_C.$$

$$4000 = 1200 + Q_C \Rightarrow Q_C = 2800 \, J.$$

Explanation: Specific heat capacity is the amount of heat required to increase the temperature of unit mass (1 kg) of a substance by 1 Kelvin (or 1 °C). It is an intensive property that depends on the material.

Explanation: Specific heat capacity measures heat per unit mass per unit temperature rise. Its SI unit is joule per kilogram per kelvin ($J/kg \, K$).

Explanation: For an ideal gas, Mayer’s relation holds:$$C_p – C_v = R,$$

where $C_p$ is specific heat at constant pressure, $C_v$ is specific heat at constant volume, and $R$ is the gas constant.

Explanation: From Mayer’s relation:$$C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R.$$

Explanation: A diatomic molecule has 5 degrees of freedom (3 translational + 2 rotational) at room temperature. By equipartition theorem:$$C_v = \frac{f}{2}R = \frac{5}{2}R.$$

At slightly higher temperature, vibrational modes become active, giving $C_v = \frac{7}{2}R$.

Explanation: Heat absorbed is given by$$Q = mc\Delta T = 2 \times 390 \times (80 – 30).$$

$$Q = 780 \times 50 = 39000 \, J = 39.0 \, kJ.$$

Explanation: The adiabatic index $\gamma$ is the ratio of molar specific heats at constant pressure and constant volume. It is used in the adiabatic process equation:$$PV^\gamma = \text{constant}.$$

Explanation: For monoatomic gases, $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$. Thus,$$\gamma = \frac{C_p}{C_v} = \frac{5/2}{3/2} = \frac{5}{3} = 1.67.$$

Explanation: For diatomic gas,$$C_v = \frac{5}{2}R, \quad C_p = C_v + R = \frac{7}{2}R.$$

$$\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4.$$

Explanation: $$Q = mc\Delta T \Rightarrow \Delta T = \frac{Q}{mc}.$$

$$\Delta T = \frac{45000}{0.5 \times 900} = \frac{45000}{450} = 100^\circ C.$$

Explanation: The basic calorimetry equation states that heat supplied to a body is proportional to its mass, specific heat capacity, and temperature rise. This relation is the foundation for measuring and calculating specific heat.

Explanation: $$Q = mc\Delta T \Rightarrow c = \frac{Q}{m\Delta T}.$$

$$c = \frac{3900}{200 \times (70 – 20)} = \frac{3900}{10000} = 0.39 \, J/gK.$$

Explanation: Heat lost by solid = Heat gained by water:$$100 \, c \, (100 – 34) = 200 \times 4.18 \times (34 – 30).$$

$$100c \times 66 = 200 \times 4.18 \times 4 = 3344.$$

$$6600c = 3344 \Rightarrow c \approx 0.21 \, J/gK.$$

Explanation: Calorimetry involves mixing a substance of known mass and temperature with water (or another reference) and measuring equilibrium temperature. This allows specific heat to be calculated from the heat balance equation.

Explanation: Heat lost by aluminum = Heat gained by water:$$250c (100 – 26) = 500 \times 4.18 (26 – 20).$$

$$250c \times 74 = 500 \times 4.18 \times 6 = 12540.$$

$$18500c = 12540 \Rightarrow c \approx 0.90 \, J/gK.$$

Explanation: To ensure accurate specific heat calculations, heat loss to surroundings must be minimized. This is achieved by using an insulated calorimeter.

Explanation: Heat lost by iron = Heat gained by water:$$1 \times c (100 – 25) = 2 \times 4186 (25 – 20).$$

$$75c = 2 \times 4186 \times 5 = 41860.$$

$$c = \frac{41860}{75} \approx 558 \, J/kgK.$$

Correction: closest known value is \~558 $J/kgK$, so option A is correct approximation.

Explanation: For one mole of a substance, molar heat capacity = specific heat × molar mass ($M$).

Explanation: Heat lost by metal = Heat gained by water:$$500c (200 – 40) = 200 \times 4.18 (40 – 25).$$

$$500c \times 160 = 200 \times 4.18 \times 15 = 12540.$$

$$80000c = 12540 \Rightarrow c \approx 0.30 \, J/gK.$$

Explanation: Calorimetry assumes that heat lost by a hot body equals heat gained by a cold body, with negligible loss to surroundings. This is a direct application of the principle of energy conservation.