### Related Formula Governing mathematical equations for thermal properties: 1. Heat capacity: C' = fracDelta QDelta T$C' = \frac{\Delta Q}{\Delta T}$ 2. Specific heat capacity: s = fracDelta Qm Delta T$s = \frac{\Delta Q}{m \Delta T}$ 3. Latent heat: L = fracDelta Qm$L = \frac{\Delta Q}{m}$ 4. Thermal conductivity: K = fracDelta Q cdot LA cdot Delta T cdot t$K = \frac{\Delta Q \cdot L}{A \cdot \Delta T \cdot t}$ ### Core Logic Let's derive the SI units for each property: - **(A) Heat capacity of body:** [C'] = fracmathrmJmathrmK = mathrmJK^-1 quad implies text(II)$[C'] = \frac{\mathrm{J}}{\mathrm{K}} = \mathrm{JK^{-1}} \quad \implies \text{(II)}$ - **(B) Specific heat capacity of body:** [s] = fracmathrmJmathrmkg cdot K = mathrmJkg^-1K^-1 quad implies text(III)$[s] = \frac{\mathrm{J}}{\mathrm{kg \cdot K}} = \mathrm{Jkg^{-1}K^{-1}} \quad \implies \text{(III)}$ - **(C) Latent heat:** [L] = fracmathrmJmathrmkg = mathrmJkg^-1 quad implies text(I)$[L] = \frac{\mathrm{J}}{\mathrm{kg}} = \mathrm{Jkg^{-1}} \quad \implies \text{(I)}$ - **(D) Thermal conductivity:** From heat transfer equation fracdQdt = K A fracdTdx$\frac{dQ}{dt} = K A \frac{dT}{dx}$: K = fracdQ/dtA (dT/dx) implies [K] = fracmathrmJ/smathrmm^2 cdot (mathrmK/m) = mathrmJ cdot m^-1 cdot K^-1 cdot s^-1 quad implies text(IV)$K = \frac{dQ/dt}{A (dT/dx)} \implies [K] = \frac{\mathrm{J/s}}{\mathrm{m}^2 \cdot (\mathrm{K/m})} = \mathrm{J \cdot m^{-1} \cdot K^{-1} \cdot s^{-1}} \quad \implies \text{(IV)}$ ### Step 1: Match values Applying our mappings: - (A) to$\to$ (II) - (B) to$\to$ (III) - (C) to$\to$ (I) - (D) to$\to$ (IV) This maps to Option (4). ### Pattern Recognition Sees: Dimensional unit matching of thermal parameters. Trap: Confusing Specific Heat Capacity (scaled per mass unit) with Heat Capacity (unscaled entire body parameter). Shortcut: Specific heat is normalized by mass, meaning its unit must contain kg in the denominator. Heat capacity has no mass constraint, meaning A maps directly to II. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula Stefan-Boltzmann Law for radiated thermal power: P = epsilon sigma A T^4$P = \epsilon \sigma A T^4$ Surface area A$A$ of a cylindrical wire of diameter d$d$ and length L$L$ is: A = pi d L$A = \pi d L$ ### Core Logic Convert given parameters to standard SI units: - L = 10 mathrm~cm = 0.1 mathrm~m$L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ - d = 0.5 mathrm~mm = 0.5 times 10^-3 mathrm~m$d = 0.5 \mathrm{~mm} = 0.5 \times 10^{-3} \mathrm{~m}$ - T = 1727 + 273.15 = 2000 mathrm~K$T = 1727 + 273.15 = 2000 \mathrm{~K}$ - P = 94.2 mathrm~W$P = 94.2 \mathrm{~W}$ - sigma = 6.0 times 10^-8 mathrm~W cdot m^-2 cdot K^-4$\sigma = 6.0 \times 10^{-8} \mathrm{~W \cdot m^{-2} \cdot K^{-4}}$ ### Step 1: Express Radiated Power and Emissivity First, calculate the surface area A$A$: A = 3.14 times (0.5 times 10^-3) times 0.1 = 1.57 times 10^-4 mathrm~m^2$A = 3.14 \times (0.5 \times 10^{-3}) \times 0.1 = 1.57 \times 10^{-4} \mathrm{~m}^2$ Substitute A$A$, sigma$\sigma$, and T$T$ into Stefan's formula: 94.2 = epsilon times (6.0 times 10^-8) times [3.14 times (0.5 times 10^-3) times (10 times 10^-2)] times (2000)^4$94.2 = \epsilon \times (6.0 \times 10^{-8}) \times [3.14 \times (0.5 \times 10^{-3}) \times (10 \times 10^{-2})] \times (2000)^4$ Calculate temperature term: (2000)^4 = 1.6 times 10^12$(2000)^4 = 1.6 \times 10^{12}$ 94.2 = epsilon times (6 times 10^-8) times (1.57 times 10^-4) times (16 times 10^12)$94.2 = \epsilon \times (6 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$ 94.2 = epsilon times 6 times 1.57 times 16 times 1 = epsilon times 150.72$94.2 = \epsilon \times 6 \times 1.57 \times 16 \times 1 = \epsilon \times 150.72$ epsilon = frac94.2150.72 = frac58$\epsilon = \frac{94.2}{150.72} = \frac{5}{8}$ Thus, fracx8 = frac58 implies x = 5$\frac{x}{8} = \frac{5}{8} \implies x = 5$. ### Pattern Recognition Sees: Bulb filament heat radiation equation. Shortcut: Simplify the multiplication with factors of 10 first. T=2000 mathrm~K$T=2000 \mathrm{~K}$ has four zeros, so T^4$T^4$ adds 10^12$10^{12}$ which cancels the 10^-8$10^{-8}$ and 10^-4$10^{-4}$ from area and constant. The coefficient equation directly yields the ratio epsilon = 5/8$\epsilon = 5/8$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula fracT_i - T_fDelta t = Kleft(fracT_i + T_f2 - T_0right)$\frac{T_i - T_f}{\Delta t} = K\left(\frac{T_i + T_f}{2} - T_0\right)$ ### Core Logic **Case 1** (90^circmathrmC rightarrow 80^circmathrmC$90^{\circ}\mathrm{C} \rightarrow 80^{\circ}\mathrm{C}$ in time t$t$): frac90 - 80t = Kleft(frac90 + 802 - 20 ight)$\frac{90 - 80}{t} = K\left(\frac{90 + 80}{2} - 20 ight)$ frac10t = K(85 - 20) = 65K quad dots (i)$\frac{10}{t} = K(85 - 20) = 65K \quad \dots (i)$ **Case 2** (80^circmathrmC rightarrow 60^circmathrmC$80^{\circ}\mathrm{C} \rightarrow 60^{\circ}\mathrm{C}$ in time t'$t'$): frac80 - 60t' = Kleft(frac80 + 602 - 20 ight)$\frac{80 - 60}{t'} = K\left(\frac{80 + 60}{2} - 20 ight)$ frac20t' = K(70 - 20) = 50K quad dots (ii)$\frac{20}{t'} = K(70 - 20) = 50K \quad \dots (ii)$ Dividing equation (i)$(i)$ by equation (ii)$(ii)$: frac10/t20/t' = frac65K50K$\frac{10/t}{20/t'} = \frac{65K}{50K}$ fract'2t = frac1310 implies t' = frac135t$\frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{5}t$ ### Pattern Recognition Newton's law of cooling in average form uses the arithmetic mean of temperatures to approximate the driving temperature difference over the interval. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula R_textth = fracLKA = fracLK(pi r^2)$R_{\text{th}} = \frac{L}{KA} = \frac{L}{K(\pi r^2)}$ [cite: 817] fracdQdt = fracDelta TR_textth$\frac{dQ}{dt} = \frac{\Delta T}{R_{\text{th}}}$ [cite: 818] ### Core Logic At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819] frac400 - TR_1 = fracT - 200R_2 implies frac400 - TT - 200 = fracR_1R_2$\frac{400 - T}{R_1} = \frac{T - 200}{R_2} \implies \frac{400 - T}{T - 200} = \frac{R_1}{R_2}$ [cite: 192, 820] Let's evaluate the resistance ratio fracR_1R_2$\frac{R_1}{R_2}$ using the dimensional parameters: [cite: 820] fracR_1R_2 = left(fracL_AL_Bright) cdot left(fracr_Br_Aright)^2 cdot left(fracK_BK_Aright)$\frac{R_1}{R_2} = \left(\frac{L_A}{L_B}\right) \cdot \left(\frac{r_B}{r_A}\right)^2 \cdot \left(\frac{K_B}{K_A}\right)$ [cite: 820] Substitute the given values: fracL_AL_B = frac12$\frac{L_A}{L_B} = \frac{1}{2}$, fracr_Ar_B = 2 implies fracr_Br_A = frac12$\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$, and fracK_AK_B = frac12 implies fracK_BK_A = 2$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$ [cite: 189, 190]: fracR_1R_2 = frac12 times left(frac12right)^2 times 2 = frac14$\frac{R_1}{R_2} = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times 2 = \frac{1}{4}$ [cite: 821] Now link this back into the temperature equation: [cite: 822] frac400 - TT - 200 = frac14 implies 1600 - 4T = T - 200$\frac{400 - T}{T - 200} = \frac{1}{4} \implies 1600 - 4T = T - 200$ [cite: 822, 823] 5T = 1800 implies T = 360\ textK$5T = 1800 \implies T = 360\ \text{K}$ [cite: 824, 825] ### Pattern Recognition Thermal conduction processes behave exactly like electric current fields in series connections[cite: 818, 819]. Cross-sectional area scales squarely with radius parameters, which requires extra care during substitution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter