A cup of coffee cools from 90^circmathrmC$90^{\circ}\mathrm{C}$ to 80^circmathrmC$80^{\circ}\mathrm{C}$ in t$t$ minutes when the room temperature is 20^circmathrmC$20^{\circ}\mathrm{C}$ . The time taken by the similar cup of coffee to cool from 80^circmathrmC$80^{\circ}\mathrm{C}$ to 60^circmathrmC$60^{\circ}\mathrm{C}$ at the same room temperature is :

### Related Formula Delta Q = m C_v Delta T$$\Delta Q = m C_v \Delta T$$ Delta A = A_0 beta Delta T = A_0 (2alpha) Delta T$$\Delta A = A_0 \beta \Delta T = A_0 (2\alpha) \Delta T$$ ### Core Logic First, calculate the temperature change using heat supplied: Delta T = fracDelta Qm C_v$$\Delta T = \frac{\Delta Q}{m C_v}$$ Substituting the values: Delta T = frac8.1 times 10^20.1 times 900 = frac81090 = 9text K$$\Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = \frac{810}{90} = 9\text{ K}$$ ### Step 1: Calculate Change in Area Initial area A_0 = ell times d = 9text cm times 4text cm = 36text cm^2 = 36 times 10^-4text m^2$A_0 = \ell \times d = 9\text{ cm} \times 4\text{ cm} = 36\text{ cm}^2 = 36 \times 10^{-4}\text{ m}^2$. Now, substitute into the area expansion formula: Delta A = 36 times 10^-4 times 2 times (3.1 times 10^-5) times 9$$\Delta A = 36 \times 10^{-4} \times 2 \times (3.1 \times 10^{-5}) \times 9$$ Delta A = 36 times 18 times 3.1 times 10^-9 = 2008.8 times 10^-9text m^2 approx 2.0 times 10^-6text m^2$$\Delta A = 36 \times 18 \times 3.1 \times 10^{-9} = 2008.8 \times 10^{-9}\text{ m}^2 \approx 2.0 \times 10^{-6}\text{ m}^2$$ ### Pattern Recognition Always remember that areal expansion coefficient beta = 2alpha$\beta = 2\alpha$. Convert geometry dimensions to standard units (1text cm^2 = 10^-4text m^2$1\text{ cm}^2 = 10^{-4}\text{ m}^2$) cleanly before concluding arithmetic. ### 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

### Related Formula fracC5 = fracF - 329$$\frac{C}{5} = \frac{F - 32}{9}$$ ### Core Logic Rearranging the conversion identity to express C$C$ as a function of F$F$: C = frac59F - frac1609$$C = \frac{5}{9}F - \frac{160}{9}$$ This is a straight line equation y = mx + c$y = mx + c$ with: - Positive slope m = frac59$m = \frac{5}{9}$ - Negative y$y$-intercept c = -frac1609$c = -\frac{160}{9}$ (at F=0$F=0$, C approx -17.8^circmathrmC$C \approx -17.8^{\circ}\mathrm{C}$) - Positive x$x$-intercept at C=0, F=32$C=0, F=32$ This completely matches the layout shown in Graph (2).

### Pattern Recognition 0^circmathrmC = 32^circmathrmF$0^{\circ}\mathrm{C} = 32^{\circ}\mathrm{F}$. Therefore, the line must cross the positive side of the horizontal F$F$ axis when C=0$C=0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula fracT_1 - T_2t = Kleft[fracT_1 + T_22 - T_s ight]$$\frac{T_1 - T_2}{t} = K\left[\frac{T_1 + T_2}{2} - T_s ight]$$ ### Core Logic For the first interval (40^circmathrmC$40^{\circ}\mathrm{C}$ to 24^circmathrmC$24^{\circ}\mathrm{C}$ in 4\ mathrmmin$4\ \mathrm{\min}$ with T_s = 16^circmathrmC$T_s = 16^{\circ}\mathrm{C}$): frac40 - 244 = Kleft[frac40 + 242 - 16 ight]$$\frac{40 - 24}{4} = K\left[\frac{40 + 24}{2} - 16 ight]$$ frac164 = K[32 - 16] implies 4 = 16K implies K = frac14$$\frac{16}{4} = K[32 - 16] \implies 4 = 16K \implies K = \frac{1}{4}$$ For the next 4 minutes, let the final temperature be T$T$: frac24 - T4 = frac14left[frac24 + T2 - 16 ight]$$\frac{24 - T}{4} = \frac{1}{4}\left[\frac{24 + T}{2} - 16 ight]$$ 24 - T = frac24 + T2 - 16$$24 - T = \frac{24 + T}{2} - 16$$ 40 - T = frac24 + T2$$40 - T = \frac{24 + T}{2}$$ 80 - 2T = 24 + T implies 3T = 56 implies T = frac563^circmathrmC$$80 - 2T = 24 + T \implies 3T = 56 \implies T = \frac{56}{3}^{\circ}\mathrm{C}$$ ### Pattern Recognition Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute K$K$ from the first step and substitute directly into the second. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Core Logic The heating curve traces temperature vs heat supplied. Stage 1: Ice at -10^circmathrmC$-10^{\circ}\mathrm{C}$ is heated to 0^circmathrmC$0^{\circ}\mathrm{C}$ (Temperature rises, solid phase). Stage 2: Ice melts at 0^circmathrmC$0^{\circ}\mathrm{C}$ into water (Temperature remains constant until all ice melts). This is a horizontal plateau. Stage 3: Water is heated from 0^circmathrmC$0^{\circ}\mathrm{C}$ to 100^circmathrmC$100^{\circ}\mathrm{C}$ (Temperature rises, liquid phase). Stage 4: Water boils at 100^circmathrmC$100^{\circ}\mathrm{C}$ into steam (Temperature remains constant). This is a second horizontal plateau. Option (4) correctly shows this sequential step-like graph. ### Pattern Recognition Heating curves always exhibit horizontal segments during phase changes (latent heat) where temperature remains constant. The slopes of the inclined regions depend on the specific heat capacities of the respective phases. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter