Apply Thermal Energy Transfers - IB Physics

1. A vacuum flask (thermos) is designed to minimize heat transfer. It consists of a double-walled glass container with a vacuum between the walls, and the surfaces facing the vacuum are silvered.
   Answer: The vacuum prevents conduction and convection; the silvering minimizes heat transfer by radiation. Heat can be transferred by conduction, convection, and radiation. A vacuum is a space devoid of particles, so it cannot support heat transfer by conduction (which requires a medium) or convection (which requires fluid movement). The silvered surfaces have low emissivity and are highly reflective. This minimizes heat transfer by radiation, both from entering the flask and from leaving it. Therefore, the vacuum addresses conduction and convection, while the silvering addresses radiation.
2. A solid substance is heated by a source of constant power. It takes 3.0 minutes to melt the substance completely at its melting point. It then takes 2.0 minutes to raise the temperature of the liquid from the melting point to the boiling point, a change of 100 K. What is the ratio of the specific latent heat of fusion ($L_f$) to the specific heat capacity of the liquid ($c_l$)?
   Answer: 150 K Let P be the power of the source. The energy supplied for melting is $Q_m = mL_f = P \times t_m$, where $t_m = 3.0$ min. The energy supplied for heating the liquid is $Q_h = mc_l\Delta T = P \times t_h$, where $t_h = 2.0$ min and $\Delta T = 100$ K. We can write the ratio $\frac{mL_f}{mc_l\Delta T} = \frac{P t_m}{P t_h}$. The mass m and power P cancel out. $\frac{L_f}{c_l\Delta T} = \frac{t_m}{t_h}$. We need to find the ratio $\frac{L_f}{c_l}$. Rearranging gives $\frac{L_f}{c_l} = \Delta T \times \frac{t_m}{t_h} = 100 \text{ K} \times \frac{3.0 \text{ min}}{2.0 \text{ min}} = 100 \times 1.5 = 150$ K.
3. An insulated container is divided into two compartments by a thermally conducting partition. One compartment contains an ideal monatomic gas X, and the other contains an ideal monatomic gas Y. The mass of a molecule of Y is four times the mass of a molecule of X ($m_Y = 4m_X$). The system is allowed to reach thermal equilibrium. What is the ratio of the root-mean-square (rms) speed of molecules in X to the rms speed of molecules in Y, $v_{rms,X} / v_{rms,Y}$?
   Answer: 2 At thermal equilibrium, both gases have the same temperature. The average translational kinetic energy of the molecules is the same for both gases: $E_{k,X} = E_{k,Y}$. The average kinetic energy is related to the rms speed by $E_k = \frac{1}{2} m v_{rms}^2$. Therefore, $\frac{1}{2} m_X v_{rms,X}^2 = \frac{1}{2} m_Y v_{rms,Y}^2$. Substituting $m_Y = 4m_X$ gives $m_X v_{rms,X}^2 = 4m_X v_{rms,Y}^2$. The mass $m_X$ cancels out, leaving $v_{rms,X}^2 = 4v_{rms,Y}^2$. Taking the square root of both sides gives $v_{rms,X} = 2v_{rms,Y}$. The ratio is $v_{rms,X} / v_{rms,Y} = 2$.
4. On a cold day, a person touches a piece of metal and a piece of wood that have both been outside for a long time. The metal feels significantly colder than the wood. What is the reason for this observation?
   Answer: The metal has a higher thermal conductivity, causing a greater rate of heat transfer from the hand. Since both objects have been outside for a long time, they are in thermal equilibrium with the surroundings and thus have the same temperature. The sensation of 'cold' is related to the rate at which heat is transferred away from the skin. Metal has a much higher thermal conductivity than wood. Therefore, it conducts heat away from the hand at a much higher rate, leading to the sensation of it being colder.
5. Consider 1 kg of liquid water at 100°C and 1 kg of steam at 100°C, both at atmospheric pressure. Which statement correctly compares their internal energies?
   Answer: The steam has greater internal energy due to the potential energy added to separate the molecules. Internal energy is the sum of the random kinetic and potential energies of the molecules. Since both substances are at the same temperature (100°C), the average kinetic energy of their molecules is the same. However, to change water into steam at the same temperature, a significant amount of energy (the latent heat of vaporization) must be added. This energy does not increase the kinetic energy but instead increases the potential energy of the molecules by overcoming the intermolecular forces and moving them farther apart. Therefore, the steam has a greater internal energy.
6. A 500 g block of an unknown metal at 250°C is placed in a well-insulated container holding 500 g of water at 20.0°C. The system reaches a final equilibrium temperature of 40.0°C. What is the specific heat capacity of the metal? (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹)
   Answer: 400 J kg⁻¹ K⁻¹ The heat lost by the metal must equal the heat gained by the water. Let $c_m$ be the specific heat capacity of the metal. $Q_{lost} = m_m c_m \Delta T_m$ and $Q_{gained} = m_w c_w \Delta T_w$. The masses are equal ($m_m = m_w = 0.5$ kg), so they will cancel. $c_m (T_{initial,m} - T_f) = c_w (T_f - T_{initial,w})$. $c_m (250 - 40) = 4200 (40.0 - 20.0)$. $c_m (210) = 4200 (20)$. $c_m = \frac{4200 \times 20}{210} = \frac{84000}{210} = 400$ J kg⁻¹ K⁻¹.
7. Two spherical stars, P and Q, radiate as black bodies. The radius of star P is twice the radius of star Q. The absolute surface temperature of star P is half the surface temperature of star Q. What is the ratio of the power radiated by P to the power radiated by Q?
   Answer: 1/4 The power radiated by a star (luminosity) is given by the Stefan-Boltzmann law, $L = \sigma A T^4$, where A is the surface area. For a sphere, $A = 4\pi r^2$. So, $L \propto r^2 T^4$. Let $r_Q$ and $T_Q$ be the radius and temperature of star Q. Then $r_P = 2r_Q$ and $T_P = 0.5T_Q$. The ratio of the powers is $\frac{L_P}{L_Q} = \frac{r_P^2 T_P^4}{r_Q^2 T_Q^4} = \frac{(2r_Q)^2 (0.5T_Q)^4}{r_Q^2 T_Q^4} = \frac{4r_Q^2 \cdot (1/16)T_Q^4}{r_Q^2 T_Q^4} = \frac{4}{16} = \frac{1}{4}$.
8. A composite rod is made of two sections of equal length and equal cross-sectional area, joined end-to-end. The thermal conductivity of material A is twice that of material B ($k_A = 2k_B$). The free end of A is held at 100°C and the free end of B is held at 0°C. What is the temperature at the junction between A and B when the rod is in a steady state of heat flow?
   Answer: 67°C In a steady state, the rate of heat flow ($P = \frac{\Delta Q}{\Delta t}$) must be the same through both sections. The formula for heat flow is $P = kA\frac{\Delta T}{x}$. Let $T_j$ be the junction temperature. For section A: $P_A = k_A A \frac{100 - T_j}{L}$. For section B: $P_B = k_B A \frac{T_j - 0}{L}$. Since $P_A = P_B$ and areas (A) and lengths (L) are equal: $k_A (100 - T_j) = k_B T_j$. Substituting $k_A = 2k_B$: $2k_B (100 - T_j) = k_B T_j$. The term $k_B$ cancels out. $2(100 - T_j) = T_j$. $200 - 2T_j = T_j$. $200 = 3T_j$. $T_j = 200/3 \approx 67$°C.
9. A deep lake with a uniform initial temperature of 8°C is cooled by cold air at its surface. What is the primary heat transfer mechanism responsible for cooling the bulk of the water down to 4°C?
   Answer: Convection, because the surface water becomes denser as it cools and sinks. Water has its maximum density at approximately 4°C. As the surface water cools from 8°C towards 4°C, it becomes denser than the warmer water below it. This denser water sinks, and the warmer, less dense water from below rises to take its place at the surface, where it is then cooled. This process creates convection currents that efficiently mix the lake water and cool the entire bulk of it towards 4°C.
10. A sealed container holds a mixture of monatomic argon gas (Ar) and diatomic nitrogen gas (N₂) in thermal equilibrium. The mass of an argon atom is greater than the mass of a nitrogen molecule. Which of the following comparisons is correct?
    Answer: The average kinetic energy of Ar atoms is the same as that of N₂ molecules. When two gases are in thermal equilibrium, they are at the same temperature (T). The average translational kinetic energy of a molecule is given by $E_k = \frac{3}{2} k_B T$, which depends only on temperature. Therefore, the average kinetic energy of the argon atoms and nitrogen molecules must be the same. Because $E_k = \frac{1}{2}mv^2$ and the masses are different, their average speeds must be different. Internal energy depends on the number of particles and degrees of freedom, which are not given, so we cannot conclude they are equal.
11. A gas undergoes an isothermal expansion at 400 K, during which its volume doubles. If the gas then undergoes an adiabatic compression that returns it to its original volume, what is the final temperature of the gas? (Assume the gas is diatomic with γ = 1.4)
    Answer: 524 K For isothermal process: $T_1 = T_2 = 400 K$, $V_2 = 2V_1$. For adiabatic compression: $T_2 V_2^{\gamma-1} = T_3 V_3^{\gamma-1}$ where $V_3 = V_1$. So $400 \times (2V_1)^{0.4} = T_3 \times V_1^{0.4}$. This gives $T_3 = 400 \times 2^{0.4} = 400 \times 1.32 = 524 K$. Choice A uses wrong γ value, B assumes isothermal throughout, C uses incorrect volume relationship.
12. A blackbody radiator has a surface temperature of 727°C. If its temperature is increased to 1227°C while maintaining the same surface area, by what factor does its total radiated power increase?
    Answer: 6.25 Stefan-Boltzmann law: $P = \sigma A T^4$. Initial temperature: $T_1 = 727 + 273 = 1000 K$. Final temperature: $T_2 = 1227 + 273 = 1500 K$. Power ratio: $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{1500}{1000}\right)^4 = (1.5)^4 = 5.06 \approx 6.25$. Choice A uses linear relationship, B uses cubic relationship, C uses squared relationship instead of fourth power.
13. A glass window has an area A, a thickness x, and a thermal conductivity k. The temperature difference between the inside and outside is $\Delta T$, resulting in a rate of heat loss P. What will be the new rate of heat loss if the area is halved, the thickness is doubled, and the temperature difference is doubled?
    Answer: P/2 The rate of heat loss by conduction is given by the formula $P = \frac{kA\Delta T}{x}$. Let the new parameters be $A' = A/2$, $x' = 2x$, and $\Delta T' = 2\Delta T$. The new rate of heat loss, $P'$, will be $P' = \frac{k A' \Delta T'}{x'} = \frac{k (A/2) (2\Delta T)}{2x} = \frac{k A \Delta T}{2x} = \frac{1}{2} \left( \frac{kA\Delta T}{x} \right) = \frac{P}{2}$.
14. A thick layer of ice has formed on a lake where the air temperature is constantly below 0°C. The water below the ice is at 0°C. Assuming a steady state, how does the rate at which the thickness of the ice increases, $\frac{dx}{dt}$, depend on the current thickness of the ice, $x$?
    Answer: The rate is inversely proportional to x. For the ice to get thicker, heat must be removed from the water at the ice-water interface and conducted through the existing ice layer to the colder air above. The rate of heat conduction is given by $P = \frac{\Delta Q}{\Delta t} = \frac{kA\Delta T}{x}$, where x is the thickness of the ice. The heat $\Delta Q$ that must be removed to freeze an additional thin layer of thickness $dx$ is $\Delta Q = L_f \rho A dx$, where $L_f$ is the latent heat of fusion and $\rho$ is the density of ice. Therefore, the rate of heat removal is $\frac{\Delta Q}{\Delta t} = L_f \rho A \frac{dx}{dt}$. Equating the two expressions for the rate of heat flow: $L_f \rho A \frac{dx}{dt} = \frac{kA\Delta T}{x}$. Rearranging for the rate of thickness increase gives $\frac{dx}{dt} = \left( \frac{k\Delta T}{L_f \rho} \right) \frac{1}{x}$. Since all terms in the parenthesis are constant, the rate $\frac{dx}{dt}$ is inversely proportional to the thickness x.
15. Which statement best describes the properties of an object that is a perfect black-body radiator?
    Answer: It absorbs all incident electromagnetic radiation and is a perfect emitter for its temperature. A perfect black body is an idealized object that absorbs all electromagnetic radiation that falls on it, regardless of frequency or angle of incidence. According to Kirchhoff's law of thermal radiation, an object that is a good absorber at a particular wavelength is also a good emitter at that wavelength. A perfect absorber (emissivity = 1) is therefore also a perfect emitter. When hot, it glows brightly (e.g., a star), so it does not appear black at all temperatures.
16. A copper block of mass 2.0 kg at 80°C is placed in thermal contact with an aluminum block of mass 1.5 kg at 20°C. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹ and of aluminum is 900 J kg⁻¹ K⁻¹. If the system reaches thermal equilibrium with negligible heat loss to the surroundings, what is the final equilibrium temperature?
    Answer: 52°C At equilibrium, heat lost by copper equals heat gained by aluminum: $m_c c_c (T_i - T_f) = m_a c_a (T_f - T_i)$. Substituting: $2.0 \times 385 \times (80 - T_f) = 1.5 \times 900 \times (T_f - 20)$. Solving: $770(80 - T_f) = 1350(T_f - 20)$, which gives $61600 - 770T_f = 1350T_f - 27000$, so $88600 = 2120T_f$, therefore $T_f = 52°C$. Choice A uses incorrect mass values, C assumes equal heat capacities, D neglects the mass difference.
17. Two solid objects, X and Y, of equal mass are supplied with thermal energy at the same constant rate. The temperature of X increases at a rate of 4.0 K s⁻¹. The temperature of Y increases at a rate of 2.0 K s⁻¹. Both objects remain in the solid phase. What is the ratio of the specific heat capacity of X to that of Y, $c_X / c_Y$?
    Answer: 0.50 The rate of energy supply, Power (P), is given by $P = \frac{dQ}{dt}$. Since $Q = mc\Delta T$, we can write $P = mc\frac{d T}{dt}$. Both objects have the same mass (m) and are supplied with energy at the same rate (P). Therefore, $m c_X (\frac{dT}{dt})_X = m c_Y (\frac{dT}{dt})_Y$. The mass m cancels out. $c_X (4.0 \text{ K s⁻¹}) = c_Y (2.0 \text{ K s⁻¹})$. Rearranging for the ratio $c_X / c_Y$ gives $\frac{c_X}{c_Y} = \frac{2.0}{4.0} = 0.50$.
18. 100 g of a liquid at 80°C is mixed with 200 g of the same liquid at 20°C in a perfectly insulated container. What is the final equilibrium temperature of the mixture?
    Answer: 40°C According to the principle of conservation of energy, the heat lost by the hot liquid equals the heat gained by the cold liquid. Let $T_f$ be the final temperature. $Q_{lost} = Q_{gained}$. $m_{hot} c (T_{hot} - T_f) = m_{cold} c (T_f - T_{cold})$. The specific heat capacity c is the same for both and cancels out. $100(80 - T_f) = 200(T_f - 20)$. Divide by 100: $1(80 - T_f) = 2(T_f - 20)$. $80 - T_f = 2T_f - 40$. $120 = 3T_f$. $T_f = 40$°C. Distractor C (50°C) is the simple average of the temperatures, which ignores the different masses.
19. A cylindrical rod expands both longitudinally and radially when heated. If the coefficient of linear expansion is 1.2 × 10⁻⁵ K⁻¹ and the temperature increases by 50°C, what is the percentage increase in the rod's volume?
    Answer: 0.180% Volume expansion coefficient $\beta = 3\alpha = 3 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-5} K^{-1}$. Fractional volume change: $\frac{\Delta V}{V_0} = \beta \Delta T = 3.6 \times 10^{-5} \times 50 = 1.8 \times 10^{-3} = 0.18\%$. Choice A uses $\alpha$ instead of $\beta$, B uses incorrect calculation, D uses $2\alpha$ instead of $3\alpha$.
20. The surface of a hot object radiates thermal energy. Due to a change in its temperature, the peak wavelength of its emitted radiation decreases by 20%. By what factor does the total power radiated per unit area from the object's surface increase?
    Answer: 2.44 According to Wien's displacement law, $\lambda_{max} T = \text{constant}$, so temperature T is inversely proportional to the peak wavelength $\lambda_{max}$. If $\lambda_{max}$ decreases by 20%, the new wavelength is $\lambda'_{max} = 0.8 \lambda_{max}$. The new temperature $T'$ is therefore $T' = T / 0.8 = 1.25T$. According to the Stefan-Boltzmann law, the power radiated per unit area is $P/A = \sigma T^4$. The new power per unit area will be proportional to $(T')^4$. The factor of increase is $\frac{(T')^4}{T^4} = (\frac{T'}{T})^4 = (1.25)^4 \approx 2.44$.

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