Welcome to Lecture 90 of our comprehensive JEE Physics Master Series. In this lecture, we conclude Unit 7 (Properties of Solids and Liquids) with an exhaustive treatment of Calorimetry, Change of State, and Heat Transfer. We examine specific heat capacities, phase equilibrium thermodynamics, Fourier’s law of thermal conduction, ice growth dynamics on freezing ponds, Newton’s law of cooling, and the governing laws of blackbody radiation (Kirchhoff, Stefan-Boltzmann, and Wien’s displacement laws).
1. Detailed Concept Card
1.1 Specific Heat Capacity, Molar Heat Capacity & Water Equivalent
Heat $Q$ is thermal energy in transit across a boundary driven by a temperature gradient. When heat is added to a system without undergoing a phase change, its temperature rises by $\Delta T$:
- Specific Heat Capacity ($s$ or $c$): The heat required to raise the temperature of a unit mass of a substance by $1\text{ K}$ (or $1^\circ\text{C}$):
$$Q = m \, s \, \Delta T \implies s = \frac{1}{m}\frac{dQ}{dT}$$
$$\text{SI Unit: } \text{J/(kg}\cdot\text{K)}, \quad \text{Practical Unit: } \text{cal/(g}\cdot^\circ\text{C)} \quad (1\text{ cal/(g}\cdot^\circ\text{C)} = 4186\text{ J/(kg}\cdot\text{K)})$$
Standard Reference Values for Water:
- Liquid Water: $s_w = 1.0\text{ cal/(g}\cdot^\circ\text{C)} = 4186\text{ J/(kg}\cdot\text{K)}$
- Ice: $s_{\text{ice}} \approx 0.50\text{ cal/(g}\cdot^\circ\text{C)} = 2100\text{ J/(kg}\cdot\text{K)}$
- Steam: $s_{\text{steam}} \approx 0.48\text{ cal/(g}\cdot^\circ\text{C)} = 2010\text{ J/(kg}\cdot\text{K)}$
- Molar Heat Capacity ($C$): Heat required per mole of substance: $$C = M_{\text{molar}} \cdot s = \frac{1}{n}\frac{dQ}{dT}$$
- Water Equivalent of a Calorimeter ($W$): The mass of water that would absorb or release the exact same quantity of heat as the calorimeter body for the same temperature change: $$W = m_{\text{cal}} \cdot \frac{s_{\text{cal}}}{s_w} = m_{\text{cal}} s_{\text{cal}} \quad [\text{in CGS where } s_w = 1\text{ cal/(g}\cdot^\circ\text{C)}]$$
1.2 Principle of Calorimetry & Latent Heat
In an isolated, thermally insulated system, the Principle of Calorimetry states that the total thermal energy remains conserved:
$$\mathbf{\sum Q_{\text{lost by hot bodies}} = \sum Q_{\text{gained by cold bodies}}}$$
Change of State & Latent Heat ($L$):
During a change of physical phase (melting, freezing, vaporization, condensation), heat is absorbed or liberated at a constant temperature. This isothermal energy is called Latent Heat:
$$Q = m \, L$$
- Latent Heat of Fusion ($L_f$): For water at $0^\circ\text{C}$: $$L_f = 80\text{ cal/g} = 3.33 \times 10^5\text{ J/kg}$$
- Latent Heat of Vaporization ($L_v$): For water at $100^\circ\text{C}$: $$L_v = 540\text{ cal/g} = 2.26 \times 10^6\text{ J/kg}$$
Phase Equilibrium Rule: In ice-water-steam mixture problems, never assume a final temperature blindly. First compute the heat required to transition all ice to water at $0^\circ\text{C}$ and compare it against the heat released by steam condensing to water at $100^\circ\text{C}$.
1.3 Thermal Conduction & Fourier’s Law
Conduction is the transfer of heat through matter via microscopic molecular collisions and free electron transport without macroscopic bulk motion of matter. Fourier’s Law of Heat Conduction states that the rate of heat flow $\frac{dQ}{dt}$ across an area $A$ is proportional to the temperature gradient $\frac{dT}{dx}$:
$$\mathbf{\frac{dQ}{dt} = -K A \frac{dT}{dx}}$$
where $K$ is the Thermal Conductivity of the material [$\text{Unit: } \text{W/(m}\cdot\text{K)}$, $\text{Dimensions: } [\text{M}^1 \text{L}^1 \text{T}^{-3} \Theta^{-1}]$].
A. The Electrical Analogy & Thermal Resistance:
In steady state through a uniform slab of length $L$ and cross-section $A$ with face temperatures $T_1$ and $T_2$ ($T_1 > T_2$):
$$H = \frac{dQ}{dt} = \frac{K A (T_1 – T_2)}{L} = \frac{\Delta T}{R_{\text{th}}}$$
The Thermal Resistance $R_{\text{th}}$ is defined as:
$$\mathbf{R_{\text{th}} = \frac{L}{K A}}$$
- Series Combination of Conductors: Heat current $H$ is constant through all slabs: $$R_{\text{eq}} = R_1 + R_2 = \frac{L_1}{K_1 A} + \frac{L_2}{K_2 A}$$ $$K_{\text{eq}} = \frac{L_1 + L_2}{\frac{L_1}{K_1} + \frac{L_2}{K_2}} \quad (\text{for } L_1 = L_2: K_{\text{eq}} = \frac{2 K_1 K_2}{K_1 + K_2})$$
- Parallel Combination of Conductors: Temperature difference $\Delta T$ is constant across all slabs: $$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \implies K_{\text{eq}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$$
B. Dynamics of Ice Layer Growth on a Freezing Lake:
Consider a sheet of ice of thickness $y$ formed on a lake where ambient air temperature is $-\theta^\circ\text{C}$ ($\theta > 0$) and the water directly below the ice is at $0^\circ\text{C}$. In time $dt$, heat conducted upward through the ice layer freezes an additional layer of thickness $dy$:
$$\frac{dQ}{dt} = \frac{K_{\text{ice}} A \theta}{y} = L_f \frac{dm}{dt} = L_f (\rho_{\text{ice}} A) \frac{dy}{dt}$$
$$y \, dy = \frac{K_{\text{ice}} \theta}{\rho_{\text{ice}} L_f} dt \implies \mathbf{t = \frac{\rho_{\text{ice}} L_f}{2 K_{\text{ice}} \theta} (y_2^2 – y_1^2)}$$
Key Ratio: The time required for ice thickness to grow from $0 \to x$, $x \to 2x$, and $2x \to 3x$ follows the ratio of differences of squares: $1^2 : (2^2 – 1^2) : (3^2 – 2^2) = \mathbf{1 : 3 : 5}$.
1.4 Convection & Newton’s Law of Cooling
Convection involves the macroscopic transfer of heat via the physical movement of warm fluid packets (natural/buoyant convection or forced convection). For a hot body at temperature $T$ cooling in ambient air at temperature $T_0$, Newton’s Law of Cooling states that the rate of heat loss is directly proportional to the temperature excess $(T – T_0)$, provided $(T – T_0) \ll T_0$:
$$\frac{dQ}{dt} = -k’ (T – T_0) \implies m s \frac{dT}{dt} = -k’ (T – T_0) \implies \mathbf{\frac{dT}{dt} = -k (T – T_0)}$$
- Exact Logarithmic Solution: $$T(t) = T_0 + (T_i – T_0) e^{-kt} \implies \ln\left(\frac{T – T_0}{T_i – T_0}\right) = -kt$$
- Approximate Mean-Temperature Method (Linear Form for Calculations): $$\frac{T_1 – T_2}{t} = k \left[\left(\frac{T_1 + T_2}{2}\right) – T_0\right]$$
1.5 Thermal Radiation & Blackbody Laws
Radiation is the transmission of thermal energy via electromagnetic waves requiring no intervening physical medium.
- Prevost’s Theory of Heat Exchange: Every body radiates energy at all temperatures above $0\text{ K}$ and simultaneously absorbs radiation from its surroundings. Thermal equilibrium represents dynamic balance, not cessation of radiation.
- Kirchhoff’s Law of Radiation: At any temperature and wavelength, the ratio of emissive power $e_\lambda$ to absorptive power $a_\lambda$ is identical for all bodies and equals the emissive power $E_\lambda$ of an ideal blackbody: $$\frac{e_\lambda}{a_\lambda} = E_\lambda \implies a_\lambda = \epsilon_\lambda$$ Takeaway: Good absorbers are inherently good emitters.
- Stefan-Boltzmann Law: The total energy radiated per second per unit surface area of an ideal blackbody is proportional to the fourth power of its absolute thermodynamic temperature $T$: $$E = \sigma T^4$$ where $\sigma = 5.670 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4)$ is Stefan’s Constant. For a real body of emissivity $\epsilon$ ($0 \le \epsilon \le 1$) and surface area $A$ in surroundings at temperature $T_0$: $$\mathbf{P_{\text{net}} = \frac{dQ}{dt} = \epsilon \, \sigma \, A (T^4 – T_0^4)}$$
- Wien’s Displacement Law: The wavelength $\lambda_{\text{max}}$ corresponding to maximum spectral radiant energy is inversely proportional to absolute temperature: $$\mathbf{\lambda_{\text{max}} \cdot T = b \approx 2.898 \times 10^{-3}\text{ m}\cdot\text{K}}$$ As temperature rises, the peak emission wavelength shifts toward higher frequencies (violet/ultraviolet).
2. Four Graded Solved Examples
Example 1: Ice-Steam Phase Equilibrium (JEE Main Level)
Problem Statement: In a thermally insulated calorimeter of negligible heat capacity, $m_1 = 10.0\text{ g}$ of ice at $-10.0^\circ\text{C}$ is mixed with $m_2 = 10.0\text{ g}$ of steam at $100.0^\circ\text{C}$. Given $s_{\text{ice}} = 0.50\text{ cal/(g}\cdot^\circ\text{C)}$, $s_w = 1.0\text{ cal/(g}\cdot^\circ\text{C)}$, $L_f = 80.0\text{ cal/g}$, and $L_v = 540.0\text{ cal/g}$, determine:
- (a) The final equilibrium temperature of the mixture.
- (b) The final mass of liquid water and steam present in the container.
Step-by-Step Solution:
Step 1: Calculate heat required to convert all ice into water at $100^\circ\text{C}$:
- Warm ice from $-10^\circ\text{C}$ to $0^\circ\text{C}$: $$Q_1 = m_1 \, s_{\text{ice}} \, \Delta T = (10.0\text{ g}) \times (0.50) \times (10^\circ\text{C}) = 50\text{ cal}$$
- Melt ice at $0^\circ\text{C}$ into water at $0^\circ\text{C}$: $$Q_2 = m_1 L_f = (10.0\text{ g}) \times (80.0\text{ cal/g}) = 800\text{ cal}$$
- Heat water from $0^\circ\text{C}$ to $100^\circ\text{C}$: $$Q_3 = m_1 \, s_w \, \Delta T = (10.0\text{ g}) \times (1.0) \times (100^\circ\text{C}) = 1000\text{ cal}$$
- Total heat required by ice: $$Q_{\text{req}} = Q_1 + Q_2 + Q_3 = 50 + 800 + 1000 = 1850\text{ cal}$$
Step 2: Calculate heat available from complete condensation of steam:
$$Q_{\text{steam, max}} = m_2 L_v = (10.0\text{ g}) \times (540.0\text{ cal/g}) = 5400\text{ cal}$$
Step 3: Analyze phase equilibrium:
Because $Q_{\text{steam, max}} = 5400\text{ cal} > Q_{\text{req}} = 1850\text{ cal}$, only a fraction of the steam needs to condense to supply the required $1850\text{ cal}$. Therefore, the mixture reaches equilibrium at $\mathbf{100.0^\circ\text{C}}$ with coexisting water and uncondensed steam.
Mass of steam that condenses into water:
$$m_{\text{condensed}} = \frac{Q_{\text{req}}}{L_v} = \frac{1850\text{ cal}}{540\text{ cal/g}} \approx 3.426\text{ g}$$
Step 4: Final composition of the mixture:
- Mass of liquid water: $m_w = m_{\text{melted ice}} + m_{\text{condensed}} = 10.0 + 3.426 = \mathbf{13.43\text{ g}}$
- Mass of steam remaining: $m_s = 10.0 – 3.426 = \mathbf{6.57\text{ g}}$
Key Takeaway: Latent heat of vaporization ($540\text{ cal/g}$) is nearly $7$ times larger than latent heat of fusion ($80\text{ cal/g}$), meaning a small mass of steam easily melts and boils an equal mass of sub-zero ice.
Example 2: Composite Thermal Conductor & Interface Temperature
Problem Statement: Two identical cylindrical rods of equal length $L = 50.0\text{ cm}$ and equal cross-sectional area $A = 10.0\text{ cm}^2$ are joined in series end-to-end. The first rod is made of aluminium ($K_1 = 200\text{ W/(m}\cdot\text{K)}$) and its exposed end is maintained at $T_1 = 100.0^\circ\text{C}$. The second rod is made of copper ($K_2 = 400\text{ W/(m}\cdot\text{K)}$) and its exposed end is maintained at $T_2 = 0.0^\circ\text{C}$. The sides are perfectly insulated. In steady state, calculate:
- (a) The junction temperature $T_j$.
- (b) The equivalent thermal conductivity $K_{\text{eq}}$ of the composite rod.
- (c) The rate of heat flow $H = \frac{dQ}{dt}$ through the composite rod.
Step-by-Step Solution:
Step 1: Calculate junction temperature $T_j$:
In steady state, the heat current flowing through both rods is identical:
$$H = \frac{K_1 A (T_1 – T_j)}{L} = \frac{K_2 A (T_j – T_2)}{L}$$
Canceling $A/L$:
$$K_1(100 – T_j) = K_2(T_j – 0) \implies 200(100 – T_j) = 400 T_j$$
$$100 – T_j = 2 T_j \implies 3 T_j = 100 \implies T_j = \frac{100}{3} \approx \mathbf{33.33^\circ\text{C}}$$
Step 2: Calculate equivalent thermal conductivity $K_{\text{eq}}$:
The total length of the composite rod is $2L$:
$$R_{\text{eq}} = R_1 + R_2 \implies \frac{2L}{K_{\text{eq}} A} = \frac{L}{K_1 A} + \frac{L}{K_2 A}$$
$$K_{\text{eq}} = \frac{2 K_1 K_2}{K_1 + K_2} = \frac{2 \times 200 \times 400}{200 + 400} = \frac{160000}{600} = \frac{800}{3} \approx \mathbf{266.67\text{ W/(m}\cdot\text{K)}}$$
Step 3: Calculate steady-state heat current $H$:
$$A = 10.0\text{ cm}^2 = 1.0 \times 10^{-3}\text{ m}^2, \quad 2L = 1.0\text{ m}$$
$$H = \frac{K_{\text{eq}} A (T_1 – T_2)}{2L} = \frac{\left(\frac{800}{3}\right) \times (1.0 \times 10^{-3}) \times (100 – 0)}{1.0} = \frac{80.0}{3} \approx \mathbf{26.67\text{ W}}$$
Key Takeaway: In thermal conduction, the junction temperature is pulled closer to the temperature of the rod with higher conductivity ($T_j = 33.3^\circ\text{C}$ is closer to $0^\circ\text{C}$ than to $100^\circ\text{C}$).
Example 3: Calculus of Ice Layer Growth on Freezing Lakes (JEE Advanced Level)
Problem Statement: The surface of a freshwater lake has frozen into an ice layer of thickness $y_1 = 2.0\text{ cm}$. The ambient atmospheric temperature above the ice is maintained steadily at $\theta = -10.0^\circ\text{C}$. The thermal conductivity of ice is $K = 2.10\text{ W/(m}\cdot\text{K)}$, its density is $\rho = 900\text{ kg/m}^3$, and latent heat of fusion is $L_f = 3.36 \times 10^5\text{ J/kg}$. Calculate:
- (a) The time $t$ (in hours) required for the ice layer to thicken from $2.0\text{ cm}$ to $4.0\text{ cm}$.
- (b) The ratio of time taken to form the second $2\text{ cm}$ layer compared to the first $2\text{ cm}$ layer ($0 \to 2\text{ cm}$).
Step-by-Step Solution:
Step 1: Set up the differential equation from Fourier’s law:
Heat released when an elemental layer of thickness $dy$ freezes over area $A$ is:
$$dQ = dm \, L_f = (\rho A \, dy) L_f$$
This heat is conducted upward through the existing ice sheet of thickness $y$ across temperature difference $\Delta T = [0^\circ\text{C} – (-10^\circ\text{C})] = 10\text{ K}$:
$$\frac{dQ}{dt} = \frac{K A \Delta T}{y} \implies \rho A L_f \frac{dy}{dt} = \frac{K A \Delta T}{y}$$
$$\int_{y_1}^{y_2} y \, dy = \frac{K \Delta T}{\rho L_f} \int_0^t dt \implies \left[ \frac{y^2}{2} \right]_{y_1}^{y_2} = \frac{K \Delta T}{\rho L_f} t$$
$$t = \frac{\rho L_f}{2 K \Delta T} (y_2^2 – y_1^2)$$
Step 2: Compute numerical constants:
$$\frac{\rho L_f}{2 K \Delta T} = \frac{900 \times (3.36 \times 10^5)}{2 \times 2.10 \times 10} = \frac{3.024 \times 10^8}{42.0} = 7.20 \times 10^6\text{ s/m}^2$$
Convert thickness limits: $y_1 = 0.02\text{ m}, y_2 = 0.04\text{ m}$:
$$y_2^2 – y_1^2 = (0.04)^2 – (0.02)^2 = 0.0016 – 0.0004 = 0.0012\text{ m}^2$$
$$t = (7.20 \times 10^6\text{ s/m}^2) \times (0.0012\text{ m}^2) = 8640\text{ seconds}$$
$$t = \frac{8640}{3600} = \mathbf{2.40\text{ hours}}$$
Step 3: Calculate the time ratio:
Time for $0 \to 2\text{ cm}$: $t_1 \propto (0.02^2 – 0) = 0.0004\text{ m}^2$.
Time for $2 \to 4\text{ cm}$: $t_2 \propto (0.04^2 – 0.02^2) = 0.0012\text{ m}^2$.
$$\frac{t_2}{t_1} = \frac{0.0012}{0.0004} = \mathbf{3.0}$$
Key Takeaway: Growing the second $2\text{ cm}$ of ice takes exactly 3 times as long as the first $2\text{ cm}$ ($7.2\text{ hr}$ vs $2.4\text{ hr}$) because existing ice acts as a self-insulating barrier that continually degrades the temperature gradient.
Example 4: Newton’s Law of Cooling vs Stefan-Boltzmann Radiation
Problem Statement: A metal sphere cools from $80.0^\circ\text{C}$ to $60.0^\circ\text{C}$ in $5.0\text{ minutes}$ in a laboratory room maintained at a constant temperature of $T_0 = 20.0^\circ\text{C}$.
- (a) Using Newton’s law of cooling in its approximate linear form, find the time $t_2$ required for the sphere to cool from $60.0^\circ\text{C}$ to $40.0^\circ\text{C}$.
- (b) Evaluate the exact time using the exponential solution.
Step-by-Step Solution:
Step 1: Approximate linear method:
For stage 1 ($80^\circ\text{C} \to 60^\circ\text{C}$):
$$\overline{T}_1 = \frac{80 + 60}{2} = 70^\circ\text{C}, \quad \Delta T_{\text{excess}} = 70 – 20 = 50^\circ\text{C}$$
$$\frac{80 – 60}{5} = k (50) \implies 4 = 50 k \implies k = 0.080\text{ min}^{-1}$$
For stage 2 ($60^\circ\text{C} \to 40^\circ\text{C}$):
$$\overline{T}_2 = \frac{60 + 40}{2} = 50^\circ\text{C}, \quad \Delta T_{\text{excess}} = 50 – 20 = 30^\circ\text{C}$$
$$\frac{60 – 40}{t_2} = k (30) \implies \frac{20}{t_2} = 0.080 \times 30 = 2.40$$
$$t_2 = \frac{20}{2.40} = \frac{25}{3} \approx \mathbf{8.33\text{ minutes}} = 8\text{ min } 20\text{ s}$$
Step 2: Exact logarithmic integration:
$$\ln\left(\frac{T – T_0}{T_i – T_0}\right) = -k t$$
Stage 1: $\ln\left(\frac{60 – 20}{80 – 20}\right) = \ln\left(\frac{40}{60}\right) = -k(5) \implies k = \frac{\ln(1.5)}{5} = \frac{0.4055}{5} = 0.0811\text{ min}^{-1}$.
Stage 2: $\ln\left(\frac{40 – 20}{60 – 20}\right) = \ln\left(\frac{20}{40}\right) = -k(t_2) \implies t_2 = \frac{\ln(2)}{k} = \frac{0.6931}{0.0811} \approx \mathbf{8.55\text{ minutes}} = 8\text{ min } 33\text{ s}$.
Key Takeaway: As the temperature difference with the surroundings drops ($50^\circ\text{C} \to 30^\circ\text{C}$), the cooling rate slows down substantially, increasing the cooling interval from $5.0\text{ min}$ to $\approx 8.3\text{ min}$.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. The molar specific heat capacity of an ideal gas during an isothermal process and an adiabatic process are respectively:
- (A) $0$ and $\infty$
- (B) $\infty$ and $0$
- (C) $0$ and $0$
- (D) $\infty$ and $\infty$
Q2. $100\text{ g}$ of ice at $0^\circ\text{C}$ is mixed with $100\text{ g}$ of water at $80^\circ\text{C}$ in a calorimeter of negligible heat capacity. The final temperature of the mixture is:
- (A) $0^\circ\text{C}$
- (B) $20^\circ\text{C}$
- (C) $40^\circ\text{C}$
- (D) $10^\circ\text{C}$
Q3. If the absolute temperature of an ideal blackbody is doubled, the total radiant power emitted per unit area increases by a factor of:
- (A) $2$
- (B) $4$
- (C) $8$
- (D) $16$
Q4. Two stars $A$ and $B$ radiate maximum energy at wavelengths $\lambda_A = 400\text{ nm}$ and $\lambda_B = 800\text{ nm}$ respectively. Assuming both stars radiate as ideal blackbodies, the ratio of total emissive power per unit area $E_A / E_B$ is:
- (A) $2$
- (B) $4$
- (C) $8$
- (D) $16$
Q5. In a steady state of heat conduction through an insulated cylindrical bar of uniform cross-section, which of the following quantities remains constant along the length of the bar?
- (A) Temperature
- (B) Temperature gradient
- (C) Rate of heat absorption
- (D) Total internal energy of elemental sections
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. Which of the following statements regarding calorimetry and change of state are correct?
- (A) During melting, the temperature remains constant while internal potential energy increases.
- (B) The water equivalent of a calorimeter has dimensions of mass $[\text{M}^1 \text{L}^0 \text{T}^0]$.
- (C) When steam at $100^\circ\text{C}$ causes a burn, it transfers more heat to the skin than water at $100^\circ\text{C}$ due to the latent heat of vaporization.
- (D) The slope of the temperature-time heating graph is directly proportional to the heat capacity of the body.
Q7. In the electrical analogy of steady-state thermal conduction through composite slabs, which of the following statements are correct?
- (A) Heat flow rate $H = \frac{dQ}{dt}$ is analogous to electric current $I$.
- (B) Temperature difference $\Delta T$ is analogous to electrical potential difference $V$.
- (C) Thermal resistance $R_{\text{th}} = \frac{L}{KA}$ has dimensions $[\text{M}^{-1} \text{L}^{-2} \text{T}^3 \Theta^1]$.
- (D) For parallel thermal conductors, the equivalent conductivity is the harmonic mean of individual conductivities.
Q8. Which of the following statements regarding thermal radiation laws are correct?
- (A) Kirchhoff’s law implies that good absorbers at a given wavelength are equally good emitters at the same wavelength.
- (B) Stefan-Boltzmann law states that total emissive power of a blackbody is proportional to $T^4$.
- (C) Wien’s displacement law states that $\lambda_{\text{max}} \propto \frac{1}{T}$.
- (D) Newton’s law of cooling is an exact consequence of Stefan’s law valid for arbitrarily large temperature differences.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A layer of ice on the surface of a pond grows from thickness $0$ to $1.0\text{ cm}$ in time $t_1$. The time taken for the ice thickness to grow further from $1.0\text{ cm}$ to $2.0\text{ cm}$ under identical atmospheric conditions is $t_2$. Calculate the ratio $t_2 / t_1$.
Q10. A spherical blackbody $A$ of radius $R_1 = 10.0\text{ cm}$ at absolute temperature $T_1$ radiates total power $P_1 = 400.0\text{ W}$. Another spherical blackbody $B$ of radius $R_2 = 20.0\text{ cm}$ is maintained at temperature $T_2 = \frac{T_1}{2}$. Calculate the total power $P_2$ (in $\text{W}$) radiated by body $B$.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (B) | Specific Heat in Thermodynamic Processes |
| Q2 | (A) | Calorimetric Ice-Water Equilibrium |
| Q3 | (D) | Stefan-Boltzmann $T^4$ Radiation Scaling |
| Q4 | (D) | Wien’s Law & Emissive Power |
| Q5 | (B) | Steady State Conduction Temperature Gradient |
| Q6 | (A), (B), (C) | Calorimetric Principles & Latent Heat |
| Q7 | (A), (B), (C) | Thermal Resistance Analogy |
| Q8 | (A), (B), (C) | Blackbody Radiation Principles |
| Q9 | 3 | Ice Layer Growth Time Ratio |
| Q10 | 100 | Blackbody Radiated Power Calculation |
Step-by-Step Solutions
Solution 1:
Specific heat capacity is defined as $C = \frac{dQ}{n \, dT}$.
- In an isothermal process, $dT = 0 \implies C = \frac{dQ}{0} = \infty$.
- In an adiabatic process, $dQ = 0 \implies C = \frac{0}{n \, dT} = 0$.
Correct Option: (B).
Solution 2:
Heat released by $100\text{ g}$ of water cooling from $80^\circ\text{C}$ to $0^\circ\text{C}$:
$$Q_{\text{lost}} = m_w s_w \Delta T = 100 \times 1 \times (80 – 0) = 8000\text{ cal}$$
Heat required to melt all $100\text{ g}$ of ice at $0^\circ\text{C}$:
$$Q_{\text{melt}} = m_{\text{ice}} L_f = 100 \times 80 = 8000\text{ cal}$$
Since $Q_{\text{lost}} = Q_{\text{melt}}$ exactly, all ice melts into water at $0^\circ\text{C}$, leaving $200\text{ g}$ of liquid water at $0^\circ\text{C}$.
Correct Option: (A).
Solution 3:
By the Stefan-Boltzmann law, $E = \sigma T^4$. When $T$ is doubled ($T’ = 2T$):
$$E’ = \sigma (2T)^4 = 16 \sigma T^4 = 16 E$$
Correct Option: (D).
Solution 4:
By Wien’s displacement law, $\lambda_{\text{max}} T = b \implies T \propto \frac{1}{\lambda_{\text{max}}}$.
$$\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A} = \frac{800\text{ nm}}{400\text{ nm}} = 2$$
By Stefan’s law:
$$\frac{E_A}{E_B} = \left(\frac{T_A}{T_B}\right)^4 = 2^4 = 16$$
Correct Option: (D).
Solution 5:
In steady state through a uniform bar with insulated sides, heat current $H = -K A \frac{dT}{dx}$ is uniform everywhere. Since $K$ and $A$ are constants, the temperature gradient $\frac{dT}{dx}$ must be constant along the entire length.
Correct Option: (B).
Solution 6:
- (A) is Correct: Latent heat breaks intermolecular lattice bonds, increasing potential energy at constant kinetic energy (constant $T$).
- (B) is Correct: $W = m_{\text{cal}} s_{\text{cal}} / s_w$ has units of grams or kilograms ($[\text{M}^1]$).
- (C) is Correct: Steam releases $540\text{ cal/g}$ of latent heat upon condensing before cooling as hot water.
- (D) is Incorrect: $\frac{dT}{dt} = \frac{1}{m s}\frac{dQ}{dt}$; the slope is inversely proportional to heat capacity $m s$.
Correct Options: (A), (B), (C).
Solution 7:
- (A) is Correct: $H = \frac{dQ}{dt} \leftrightarrow I = \frac{dq}{dt}$.
- (B) is Correct: $\Delta T \leftrightarrow \Delta V$.
- (C) is Correct: $[R_{\text{th}}] = \frac{[\Theta]}{[\text{M}^1 \text{L}^2 \text{T}^{-3}]} = [\text{M}^{-1} \text{L}^{-2} \text{T}^3 \Theta^1]$.
- (D) is Incorrect: For parallel conductors, $K_{\text{eq}}$ is the weighted arithmetic mean ($K_{\text{eq}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$), not the harmonic mean.
Correct Options: (A), (B), (C).
Solution 8:
- (A) is Correct: Kirchhoff’s law $e_\lambda / a_\lambda = E_\lambda$ implies $a_\lambda = \epsilon_\lambda$.
- (B) is Correct: $E = \sigma T^4$.
- (C) is Correct: $\lambda_{\text{max}} T = \text{constant}$.
- (D) is Incorrect: Newton’s law of cooling is an approximation obtained by linearizing $(T^4 – T_0^4) \approx 4 T_0^3 (T – T_0)$, valid only when $(T – T_0) \ll T_0$.
Correct Options: (A), (B), (C).
Solution 9:
From the ice growth formula $t \propto (y_2^2 – y_1^2)$:
$$t_1 \propto (1.0^2 – 0^2) = 1.0$$
$$t_2 \propto (2.0^2 – 1.0^2) = 4.0 – 1.0 = 3.0$$
$$\frac{t_2}{t_1} = \frac{3.0}{1.0} = 3$$
Final Answer: 3
Solution 10:
Total radiated power is $P = 4\pi R^2 \sigma T^4$:
$$\frac{P_2}{P_1} = \left(\frac{R_2}{R_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{20}{10}\right)^2 \left(\frac{1}{2}\right)^4 = (2)^2 \times \frac{1}{16} = 4 \times \frac{1}{16} = \frac{1}{4}$$
$$P_2 = \frac{P_1}{4} = \frac{400.0\text{ W}}{4} = 100\text{ W}$$
Final Answer: 100