Match List-I with List-II. List-I: (A) Heat capacity of body (B) Specific heat capacity of body (C) Latent heat (D) Thermal conductivity List-II: (I) mathrmJkg^-1 (II) mathrmJK^-1 (III) mathrmJkg^-1K^-1 (IV) mathrmJm^-1mathrmK^-1mathrms^-1 Choose the correct answer from the options given below:

Solution & Explanation

### Related Formula Governing mathematical equations for thermal properties: 1. Heat capacity: C' = fracDelta QDelta T 2. Specific heat capacity: s = fracDelta Qm Delta T 3. Latent heat: L = fracDelta Qm 4. Thermal conductivity: K = fracDelta Q cdot LA 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) - **(B) Specific heat capacity of body:** [s] = fracmathrmJmathrmkg cdot K = mathrmJkg^-1K^-1 quad implies text(III) - **(C) Latent heat:** [L] = fracmathrmJmathrmkg = mathrmJkg^-1 quad implies text(I) - **(D) Thermal conductivity:** From heat transfer equation fracdQdt = K A fracdTdx: 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) ### Step 1: Match values Applying our mappings: - (A) to (II) - (B) to (III) - (C) to (I) - (D) 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

Reference Study Guides

More Thermal Properties of Matter Previous-Year Questions

Q27 jee_main_2026_21_jan_morning Calorimetry
A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27^circtextC to 87^circtextC. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 times 10^4 J/g, specific heat capacity of water = 4200text J/kg.^circtextC)
  • A. 2.1
  • B. 4.2
  • C. 0.42
  • D. 0.21

Solution

### Related Formula P = fracdmdt S Delta T P_textheater = textRate of combustion times textHeat of combustion ### Core Logic Water flow rate = 5text l/min = frac560text kg/s = frac112text kg/s. The power required to heat the water is: P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W ### Step 1: Solving for Gas Consumption Let the rate of consumption of gas be xtext g/s. Heat generated by the gas per second must equal the power required to heat the water: x times 5.0 times 10^4 = frac112 times 4200 times 60 x times 5.0 times 10^4 = 4200 times 5 x = frac4200 times 55 times 10^4 = 4200 times 10^-4 x = 0.42text g/s
Calorimetry diagram for Q27 - JEE Main 2026 Morning
Calorimetry diagram for Q27 - JEE Main 2026 Morning
### Pattern Recognition Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T) with the power output of fuel (dm_textfuel/dt cdot L_textcombustion). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter
Q40 jee_main_2026_21_jan_morning Thermal Expansion
An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30°C. The coefficient of linear expansion of aluminium and steel are 24 times 10^-6/^circmathrmC and 1.2 times 10^-5/^circmathrmC, respectively. The length of this composite rod when its temperature is raised to 100°C, is ____ cm.
  • A. 120.20
  • B. 120.15
  • C. 120.03
  • D. 120.06

Solution

### Related Formula Delta l = l_0 alpha Delta T l_textfinal = l_1 + l_2 + Delta l_1 + Delta l_2 ### Core Logic Let the original lengths be l_0A (aluminium) and l_0S (steel). They have same length, so l_0A = l_0S = 60text cm. Change in temperature Delta T = 100^circtextC - 30^circtextC = 70^circtextC. Expansion of Aluminium: Delta l_A = l_0 alpha_A Delta T Expansion of Steel: Delta l_S = l_0 alpha_S Delta T ### Step 1: Calculate Final Length l_textfinal = l_0(1 + alpha_A Delta T) + l_0(1 + alpha_S Delta T) l_textfinal = l_0 [2 + (alpha_A + alpha_S)Delta T] l_textfinal = 60 left[ 2 + (24 times 10^-6 + 12 times 10^-6) times 70 right] l_textfinal = 60 left[ 2 + (36 times 10^-6) times 70 right] l_textfinal = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15text cm
Thermal Expansion diagram for Q40 - JEE Main 2026 Morning
Thermal Expansion diagram for Q40 - JEE Main 2026 Morning
### Pattern Recognition For composite rods connected end-to-end, expansions simply add up. Convert 1.2 times 10^-5 to 12 times 10^-6 immediately for easy mental addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter
Q24 jee_main_2025_07_april_morning Radiation
A wire of length 10mathrmcm and diameter 0.5mathrmmm is used in a bulb. The temperature of the wire is 1727^circmathrmC and power radiated by the wire is 94.2 W. Its emissivity is fracmathrmx8 where mathrmx = (Given sigma = 6.0 times 10^-8 mathrm~W mathrm~m^-2 mathrm~K^-4 , pi = 3.14 and assume that the emissivity of wire material is same at all wavelength.)
Numerical Answer. Answer: 5 to 5

Solution

### Related Formula Stefan-Boltzmann Law for radiated thermal power: P = epsilon sigma A T^4 Surface area A of a cylindrical wire of diameter d and length L is: A = pi d L ### Core Logic Convert given parameters to standard SI units: - L = 10 mathrm~cm = 0.1 mathrm~m - d = 0.5 mathrm~mm = 0.5 times 10^-3 mathrm~m - T = 1727 + 273.15 = 2000 mathrm~K - P = 94.2 mathrm~W - 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 = 3.14 times (0.5 times 10^-3) times 0.1 = 1.57 times 10^-4 mathrm~m^2 Substitute A, sigma, and 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 Calculate temperature term: (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 1.57 times 16 times 1 = epsilon times 150.72 epsilon = frac94.2150.72 = frac58 Thus, fracx8 = frac58 implies x = 5. ### Pattern Recognition Sees: Bulb filament heat radiation equation. Shortcut: Simplify the multiplication with factors of 10 first. T=2000 mathrm~K has four zeros, so T^4 adds 10^12 which cancels the 10^-8 and 10^-4 from area and constant. The coefficient equation directly yields the ratio epsilon = 5/8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter
Q19 jee_main_2025_29_jan_evening Newton's Law of Cooling
A cup of coffee cools from 90^circmathrmC to 80^circmathrmC in t minutes when the room temperature is 20^circmathrmC . The time taken by the similar cup of coffee to cool from 80^circmathrmC to 60^circmathrmC at the same room temperature is :
  • A. frac135 t
  • B. frac1013 t
  • C. frac1310 t
  • D. frac513 t

Solution

### Related Formula fracT_i - T_fDelta t = Kleft(fracT_i + T_f2 - T_0right) ### Core Logic **Case 1** (90^circmathrmC rightarrow 80^circmathrmC in time t): frac90 - 80t = Kleft(frac90 + 802 - 20 ight) frac10t = K(85 - 20) = 65K quad dots (i) **Case 2** (80^circmathrmC rightarrow 60^circmathrmC in time t'): frac80 - 60t' = Kleft(frac80 + 602 - 20 ight) frac20t' = K(70 - 20) = 50K quad dots (ii) Dividing equation (i) by equation (ii): frac10/t20/t' = frac65K50K fract'2t = frac1310 implies t' = frac135t ### 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

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