Physics · Properties of bulk matter · chapter 10 of 14

Class 11 Physics Chapter 10: Thermal Properties of Matter worksheet with answer key

A 20-mark worksheet on Thermal Properties of Matter for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

A beaker of pure water at 0 °C is warmed slowly to 8 °C. Over this range its density

(a) increases throughout(b) decreases throughout(c) first decreases, then increases(d) first increases, then decreases

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Answer: (d) first increases, then decreases

Water contracts on heating from 0 °C to 4 °C, so its density rises; above 4 °C it expands and the density falls. Water is densest at 4 °C.

Question 2

Rods of equal length made of iron (α_l = 1.2 × 10⁻⁵ K⁻¹), aluminium (2.5 × 10⁻⁵ K⁻¹), copper (1.7 × 10⁻⁵ K⁻¹) and brass (1.8 × 10⁻⁵ K⁻¹) are heated through the same rise in temperature. The rod that lengthens the most is the one made of

(a) iron(b) aluminium(c) copper(d) brass

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Answer: (b) aluminium

Δl = α_l l ΔT, so with l and ΔT the same, the rod with the largest α_l lengthens the most: aluminium, 2.5 × 10⁻⁵ K⁻¹.

Question 3

An aluminium rod 2.0 m long is heated from 20 °C to 120 °C (α_l = 2.5 × 10⁻⁵ K⁻¹). Its length increases by

(a) 0.5 mm(b) 2.5 mm(c) 5.0 mm(d) 50 mm

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Answer: (c) 5.0 mm

Δl = α_l l ΔT = 2.5 × 10⁻⁵ × 2.0 × 100 = 5.0 × 10⁻³ m = 5.0 mm. 2.5 mm forgets the length of 2.0 m; 0.5 mm and 50 mm slip a power of ten.

Question 4

An ideal gas is heated at constant pressure. At 27 °C its coefficient of volume expansion is about

(a) 3.7 × 10⁻³ K⁻¹(b) 3.7 × 10⁻² K⁻¹(c) 1.1 × 10⁻³ K⁻¹(d) 3.3 × 10⁻³ K⁻¹

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Answer: (d) 3.3 × 10⁻³ K⁻¹

For an ideal gas at constant pressure α_v = 1/T, with T in kelvin: 1/300 K = 3.3 × 10⁻³ K⁻¹. 3.7 × 10⁻³ K⁻¹ is the value at 0 °C; 3.7 × 10⁻² K⁻¹ uses 1/27, the Celsius number.

Question 5

The temperature of a room rises from 18 °C to 43 °C. The rise in temperature on the Kelvin scale is

(a) 25 K(b) 298.15 K(c) 45 K(d) 316.15 K

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Answer: (a) 25 K

The Kelvin and Celsius scales have the same size of unit (T = t_C + 273.15), so a rise of 25 °C is a rise of 25 K. 298.15 K and 316.15 K are temperatures, not a rise; 45 is the rise in °F.

Question 6 · 2 marks

Equal volumes of ethanol and mercury are warmed through the same rise in temperature. With α_v(ethanol) = 110 × 10⁻⁵ K⁻¹ and α_v(mercury) = 18.2 × 10⁻⁵ K⁻¹, find the ratio of their increases in volume. In two thermometers with identical bulbs and bores, which liquid gives the longer length per degree?

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ΔV = α_v VΔT, so with V and ΔT equal, ΔV_ethanol/ΔV_mercury = 110/18.2 ≈ 6.0. Ethanol expands about six times as much, so its column moves farther for each degree: the ethanol thermometer has the longer length per degree.

ratio ≈ 6.0 from ΔV = α_v VΔT (1)
ethanol, with reason (1)

Question 7 · 2 marks

Name two features of a Dewar flask (thermos bottle) and, for each, the mode of heat transfer it reduces.

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The inner and outer walls are coated with silver: radiation from the inner wall is reflected back to the contents and incoming radiation is reflected back by the outer wall, which reduces heat transfer by radiation. The space between the double walls is evacuated, which reduces conduction and convection. (The flask also rests on an insulator such as cork, reducing conduction.)

silvered walls: radiation (1)
evacuated gap: conduction and convection (1)

Question 8 · 2 marks

What is sublimation? Name two substances that sublime at normal atmospheric pressure, and state how the solid and vapour are related while the change takes place.

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Sublimation is the change from the solid state directly to the vapour state without passing through the liquid state. Dry ice (solid CO₂) and iodine sublime. During sublimation the solid and the vapour coexist in thermal equilibrium.

definition of sublimation (1)
two examples and solid-vapour equilibrium (1)

Question 9 · 3 marks

A hot plate of surface area 0.050 m² and emissivity 0.80 is kept at 500 K by an electric heater, in a room whose walls are at 300 K. Radiation is the only loss. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴) (a) Find the power the plate radiates. (b) Find the heater power needed to keep the plate at 500 K. (c) By what factor must the heater power rise to keep the plate at 600 K instead?

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(a) H = eσAT⁴ = 0.80 × 5.67 × 10⁻⁸ × 0.050 × (500)⁴ = 2.268 × 10⁻⁹ × 6.25 × 10¹⁰ ≈ 141.8 W. (b) The plate also absorbs from the walls, so the heater supplies the net loss: eσA(T⁴ − T_s⁴) = 2.268 × 10⁻⁹ × (6.25 × 10¹⁰ − 8.1 × 10⁹) ≈ 123.4 W. (c) (600⁴ − 300⁴)/(500⁴ − 300⁴) = (12.96 − 0.81)/(6.25 − 0.81) ≈ 2.23.

radiated power ≈ 141.8 W (1)
net loss ≈ 123.4 W = heater power (1)
factor ≈ 2.23 (1)

Question 10 · 3 marks

A copper rod and a pyrex glass rod are each 1.000 m long at 20 °C. Both are heated to 120 °C. With α_l(copper) = 1.7 × 10⁻⁵ K⁻¹ and α_l(pyrex) = 0.32 × 10⁻⁵ K⁻¹, find (a) the increase in length of each rod and (b) how many times the copper's increase is the glass's. (c) State why pyrex glassware is preferred for heating in a laboratory.

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(a) Δl = α_l lΔT. Copper: 1.7 × 10⁻⁵ × 1.000 × 100 = 1.7 × 10⁻³ m = 1.7 mm. Pyrex: 0.32 × 10⁻⁵ × 1.000 × 100 = 0.32 × 10⁻³ m = 0.32 mm. (b) 1.7/0.32 ≈ 5.3: copper expands about five times as much. (c) Pyrex has a very small α_l, so uneven heating sets up little difference in expansion across the glass and it does not crack.

both increases: 1.7 mm and 0.32 mm (1)
ratio ≈ 5.3 (1)
small α_l, little uneven expansion (1)

Question 11 · 3 marks

0.50 kg of a solid is heated by a 200 W heater, all of whose heat goes into the substance. Its temperature rises, then stays at 80 °C for 4.0 minutes, and then rises again, the liquid warming by 20 °C in 50 s. (a) What is happening during the 4.0 minutes? (b) Find the latent heat of fusion of the substance. (c) Find the specific heat capacity of the liquid.

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(a) The substance is melting at its melting point, 80 °C; the heat supplied changes its state, solid and liquid coexisting, so the temperature does not rise. (b) Q = 200 × 240 = 48000 J; L_f = Q/m = 48000/0.50 = 96000 J kg⁻¹ = 9.6 × 10⁴ J kg⁻¹. (c) Q = 200 × 50 = 10000 J; s = Q/(mΔT) = 10000/(0.50 × 20) = 1000 J kg⁻¹ K⁻¹.

melting at constant temperature (1)
L_f = 9.6 × 10⁴ J kg⁻¹ (1)
s = 1000 J kg⁻¹ K⁻¹ (1)

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