Thermal Properties of Matter — Practice

Question 1

Convert a temperature of 350 K to Celsius and Fahrenheit.

Accepted Answer

t_C = 350 - 273.15 = 76.85 C. t_F = (9/5)(76.85) + 32 = 138.33 + 32 = 170.33 F. Always subtract 273.15, not the rounded 273, for the exact conversion.

Question 2

Why does the anomalous expansion of water between 0C and 4C allow fish to survive in a frozen pond?

Accepted Answer

As the pond cools toward 4C, denser surface water sinks and warmer water rises, mixing normally. Once the surface drops below 4C it becomes LESS dense than the water beneath it, so it stays on top and freezes there first, insulating the liquid water below. Without this anomalous behaviour, a pond would freeze from the bottom up, and nothing could survive it.

Question 3

State Kirchhoff's law of radiation and use it to explain why a shiny metal teapot stays hotter for longer than a matte black one at the same starting temperature.

Accepted Answer

Kirchhoff's law: a good absorber of radiation at a given temperature is equally a good emitter of it, and a poor absorber is equally a poor emitter. A shiny surface reflects most radiation and absorbs little, so by the law it also emits little, radiating heat away slowly and staying hot longer. The matte black teapot, a good absorber, is equally a good emitter, and cools faster.

Question 4

A steel rod is 2.000 m long at 20C. Find its length at 120C. Coefficient of linear expansion of steel = 1.2 x 10^-5 K^-1.

Accepted Answer

dL = L0 alpha dT = 2.000 x 1.2e-5 x 100 = 2.4e-3 m = 2.4 mm. New length = 2.000 + 0.0024 = 2.0024 m.

Question 5

200 g of water at 80C is mixed with 300 g of water at 20C. Find the final temperature, assuming no heat loss.

Accepted Answer

Heat lost by hot water = heat gained by cold water: 200 x s x (80-Tf) = 300 x s x (Tf-20). The specific heat cancels since both are water: 200(80-Tf) = 300(Tf-20). 16000-200Tf = 300Tf-6000. 22000 = 500Tf. Tf = 44 C.

Question 6

Find the heat required to convert 50 g of ice at 0C completely into water at 0C. Latent heat of fusion of ice = 335 J/g.

Accepted Answer

Q = mL = 50 x 335 = 16,750 J. The temperature does not change at all during this process — all the heat goes into breaking the bonds that hold the solid structure together.

Question 7

A metal rod of length 2.0 m and cross-sectional area 4.0 cm^2 conducts heat at a rate of 20 W when its ends are kept at 100C and 0C. Find its thermal conductivity.

Accepted Answer

H = KA(dT)/d, so K = Hd/(A dT) = (20 x 2.0)/(4.0e-4 x 100) = 40/0.04 = 1000 W/mK. This is an unusually high value, closer to a very good conductor like silver, which is worth noting as a sanity check on any answer of this size.

Question 8

A body cools from 90C to 60C in 10 minutes when the surroundings are at 20C. Find its temperature after a further 10 minutes.

Accepted Answer

Using the exponential form: (60-20)/(90-20) = e^(-10k), so 40/70 = e^(-10k), giving k = ln(7/4)/10 = 0.0560/min. After a further 10 minutes (t=20 min total): (Tf-20)/(90-20) = e^(-20k) = e^(-1.120) = 0.3265. Tf = 20 + 70(0.3265) = 20+22.85 = 42.9 C.

Question 9

A brass rod and a steel rod, each 1.0 m long at 20C, are clamped rigidly end to end between two fixed rigid supports, with a total gap of exactly 2.0 m between the supports at 20C. Find the compressive stress in each rod when the temperature rises to 100C, given both rods have the same cross-sectional area, alpha_brass = 2.0e-5/K, alpha_steel = 1.2e-5/K, Y_brass = 0.91e11 Pa, Y_steel = 2.0e11 Pa.

Accepted Answer

Since the total gap is fixed, the combined free expansion of both rods must be entirely converted into compressive strain, shared between them so that both press with the SAME force F (they are in series between rigid supports). Free expansion: dL_brass=1.0x2.0e-5x80=1.6e-3 m, dL_steel=1.0x1.2e-5x80=0.96e-3 m, total=2.56e-3 m. This must be absorbed as compression: F/A x [L/Y_brass + L/Y_steel] = total dL, so F/A = 2.56e-3 / [1.0/0.91e11 + 1.0/2.0e11] = 2.56e-3/(1.099e-11+0.500e-11) = 2.56e-3/1.599e-11 = 1.601e8 Pa. Since the rods are in series carrying the same force, both rods experience the SAME compressive stress, about 1.6 x 10^8 Pa.

Question 10

10 g of steam at 100C is passed into 100 g of water at 24C until the temperature of the water rises to 40C. Find the mass of steam condensed. Latent heat of steam = 2256 J/g, specific heat of water = 4.186 J/gC.

Accepted Answer

Let m grams of steam condense. Heat released = m x 2256 (condensing) + m x 4.186 x (100-40) (cooling the condensed water from 100 to 40) = m(2256+251.2) = 2507.2m. Heat absorbed by the 100g of water warming from 24 to 40: Q = 100 x 4.186 x 16 = 6697.6 J. Setting released equal to absorbed: 2507.2m = 6697.6, so m = 2.67 g. Only a small mass of steam condenses because its latent heat is so large compared with the heat needed to warm the water.

Practice — Thermal Properties of Matter

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