Previous year questions: Thermal Properties of Matter

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the formulas on the first page of this chapter, and every answer is worked out in full so that the method behind the correct option is clear.

Q1 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

A metal strip has coefficient of linear expansion alpha. Its temperature is raised from T1 to T2, and its length increases by delta L1. The temperature is then raised further from T2 to T3, and the length increases by delta L2. Given T3 + T1 = 2 T2 and T2 - T1 = delta T, find delta L2.

Answer: The first increase is delta L1 = L alpha delta T. Also T3 - T2 = T2 - T1 = delta T, and the second expansion starts from the already expanded length L + delta L1, so delta L2 = (L + delta L1) alpha delta T = delta L1 + alpha delta L1 delta T = delta L1 (1 + alpha delta T). The option delta L1 (1 + alpha delta T) is correct. The second expansion starts from the already expanded length, which is where the extra alpha delta T factor comes from.

Q2 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

The heat taken out of x gram of water, initially at 50 degrees C, to cool it to 0 degrees C is enough to evaporate (1000 - x) gram of water that is also initially at 50 degrees C. Find x, to the nearest integer. Take the latent heat of water as 2256 kJ kg^-1 and the specific heat capacity of water as 4200 J kg^-1 K^-1.

Answer: Cooling x g of water from 50 to 0 releases x c (50), where c is the specific heat in J kg^-1 K^-1. Evaporating the other part first heats it from 50 to 100 degrees C and then vaporises it, so the heat needed is (1000 - x)(c(50) + L) in grams, with L converted to J kg^-1. In grams and J, the equation is x(4200 x 50) = (1000 - x)(4200 x 50 + 2256 x 10^3). This reduces to 210000x = 2466000(1000 - x), and x = 2466000000/2676000 = 921.5, so x is about 922 g.

Q3 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

An ideal gas at pressure P and temperature T expands in such a way that P T^3 stays constant. Find the coefficient of volume expansion of the gas.

Answer: Write P = C/T^3 and use the ideal gas equation V = nRT/P = nRT^4/C. The coefficient of volume expansion is gamma = (1/V) dV/dT at constant pressure. Differentiating V gives dV/dT = 4nRT^3/C, so gamma = 4nRT^3/(C) divided by nRT^4/C = 4/T. The option 4/T is correct.

Q4 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Match the constants in List I with their dimensions in List II: A. Boltzmann constant, B. Stefan's constant, C. Planck's constant, D. gravitational constant, with I [M^-1 L^3 T^-2], II [M L^2 T^-1], III [M L^2 T^-2 K^-1] and IV [M L^0 T^-3 K^-4].

Answer: Boltzmann constant has energy per kelvin, so its dimension is [M L^2 T^-2 K^-1], which is III. Planck's constant has the dimension of angular momentum, [M L^2 T^-1], which is II. The gravitational constant follows from F = Gm1m2/r^2, giving [M^-1 L^3 T^-2], which is I. Stefan's constant has dimension [M L^0 T^-3 K^-4], IV. So the matching is A-III, B-IV, C-II, D-I.

Q5 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

Five moles of an unknown gas is heated at constant volume from 10 degrees C to 20 degrees C. The molar specific heat of the gas at constant pressure is c_p = 8 cal mol^-1 K^-1, and R = 8.36 J mol^-1 K^-1. Find the change in the internal energy of the gas in calorie.

Answer: At constant volume the heat supplied equals the change in internal energy, and it is calculated with C_v. Using C_v = C_p - R = 8 x 4.186 - 8.36 = 33.49 - 8.36 = 25.13 J mol^-1 K^-1, the change is delta U = n C_v delta T = 5 x 25.13 x 10 = 1256 J. Converting to calories with 1 cal = 4.186 J gives 300 cal. The answer is 300.

Q6 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

Two closed vessels of equal volume V are joined by a narrow tube. Both contain air at 90 kPa and 400 K. The temperature of one vessel is held at 400 K while the other is raised to 500 K. Find the final pressure in the vessels.

Answer: Let the final pressure be P. Initially each vessel holds 90 kPa at 400 K, so the total number of moles is 2 x (90 x 10^3 V)/(R x 400). Finally one vessel is at 400 K and the other at 500 K, and both share the pressure P, so the total number of moles is P V/(R x 400) + P V/(R x 500). Equating the two and cancelling V/R gives P(1/400 + 1/500) = 2 x 90/400, which is P x 4.5 x 10^-3 = 0.45 when P is in kPa, so P = 100 kPa. The option 100 is correct. The same reasoning applies whenever two parts of a gas share one pressure at different temperatures.

Q7 (NCERT Exemplar, 1 mark)

A bimetallic strip is made of aluminium and steel, with the coefficient of linear expansion of aluminium greater than that of steel. On heating, how does the strip behave?

Answer: Aluminium expands more than steel for the same rise in temperature, so the aluminium side becomes the longer arc of the bend and lies on the outside, that is on the convex side. The steel side lies on the inside, that is the concave side. The correct choice is that the strip bends with steel on the concave side. A thermostat uses that bend to switch a heater or a cooler on and off as the temperature changes.

Q8 (NCERT Exemplar, 1 mark)

At what temperature does the numerical value on the Celsius scale equal the numerical value on the Fahrenheit scale?

Answer: Set t_C = t_F = x in t_F = (9/5)t_C + 32. Then x = 1.8x + 32, so -0.8x = 32 and x = -40. The two scales read the same number at -40 degrees, which is the only such point. The same equation shows why the two scales cannot agree anywhere else.

Q9 (NCERT Exemplar, 3 marks)

100 g of water is supercooled to -10 degrees C. Some of it suddenly freezes to ice after a disturbance. Find the final temperature of the mixture and the mass that freezes. Take the specific heat of water as 1 cal g^-1 C^-1 and the latent heat of fusion as 80 cal g^-1.

Answer: The heat needed to warm all 100 g of water from -10 to 0 degrees C is 100 x 1 x 10 = 1000 cal. This heat comes from freezing part of the water, so the mass that freezes is m = 1000/80 = 12.5 g. The rest of the water stays liquid at 0 degrees C, so the final temperature of the mixture is 0 degrees C, with 87.5 g liquid and 12.5 g ice.

Q10 (NCERT Exemplar, 3 marks)

A tooth cavity is filled with copper and hot tea at 57 degrees C is drunk. Taking the tooth temperature as 37 degrees C, the coefficient of linear expansion of copper as 1.7 x 10^-5 per degree C and the bulk modulus of copper as 1.4 x 10^11 N m^-2, estimate the thermal stress developed in the copper.

Answer: The copper would try to expand by a volume strain of 3 alpha delta T = 3 x 1.7 x 10^-5 x 20 = 1.02 x 10^-3. With the cavity preventing the expansion, the stress is the bulk modulus times that strain, so stress = 1.4 x 10^11 x 1.02 x 10^-3 = 1.43 x 10^8 N m^-2. The estimate is of the order of 10^8 Pa, which is why a sudden hot drink can damage a filled tooth. The same calculation explains why dental fillings that match the expansion coefficient of the tooth material are preferred.