Solution Tutorial 1

SOLUTION TUTORIAL 1

1.

A piece of copper is dropped into a beaker of water. If the water’s temperature rises, what happens to the temperature of copper? Under what conditions are the water and copper in thermal equilibrium? Solution: The copper’s temperature drops and the water temperature rises until both temperatures are the same. Then the metal and the water are in thermal equilibrium.

2.

Markings to indicate length are placed on a steel tape in a room that has a temperature of 22°C. Are measurements made with the tape on a day when the temperature is 27°C too long, too short, or accurate? Defend your answer. Solution: The measurements made with the heated steel tape will be too short—but only by a factor −5 of 5 × 10 of the measured length.

3.

Solution: The sphere expands when heated, so that it no longer fits through the ring. With the sphere still hot, you can separate the sphere and ring by heating the ring. This more surprising result occurs because the thermal expansion of the ring is not like the inflation of a blood-pressure cuff. Rather, it is like a photographic enlargement; every linear dimension, including the hole diameter, increases by the same factor. The reason for this is that the atoms everywhere, including those around the inner circumference, push away from each other. The only way that the atoms can accommodate the greater distances is for the circumference—and corresponding diameter—to grow. This property was once used to fit metal rims to wooden wagon and horse-buggy wheels. If the ring is heated and the sphere left at room temperature, the sphere would pass through the ring with more space to spare.

4.

Metal lids on a glass jars can often be loosed by running hot water over them. How is that possible? Solution: The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar, both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a certain temperature at which the inner diameter of the lid has expanded more than the top of the jar, and the lid will be easier to remove.

5.

Liquid nitrogen has a boiling point of –195.81°C at atmospheric pressure. Express this temperature in (a) degrees Fahrenheit and (b) Kelvin Solution: (a)

(b)

TF =

9 9 TC + 32.0°F = (−195.81) + 32.0 = −320°F 5 5

T = TC + 273.15 = −195.81 + 273.15 = 77.3 K

6.

What are the following temperature on the Kelvin scale: a) 86 ºC b) 78 ºF c) -100 ºC d) – 459 ºF Solution: a) 86 ºC = 359 K b) 78 ºF = 299 K c) -100 ºC = 173 K d) – 459 ºF = 0.37 K

7.

What would be the final temperature of a mixture of 50 g of 20 ºC water and 50 g of 40 ºC water?

Solution: Heat gained by the cooler water = heat lost by the warmer water mc∆θ = mc∆θ (θ - 20º) = (40º - θ) θ = 30º

8.

How much heat is required to raised the temperature of 250 ml of water from 20.0 ºC to 35.0 ºC. How much heat is lost by the water as it cools back down to 20.0 ºC? (Given, c = 1.00 cal/g. ºC)

Solution: (a) Q = mc ∆T = (250 g) (1.00 cal/g. ºC) (15.0 ºC) = 3.75 x 103 cal = 15.7 kJ (b) Q = mc ∆T = (250 g) (1.00 cal/g. ºC) (-15.0 ºC) = - 3.75 x 103 cal = - 15.7 kJ

9.

A 1.50-kg iron horseshoe initially at 600ºC is dropped into a bucket containing 20.0-kg of water at 25 ºC. What is the final temperature? (cw = 4186 J/kg. ºC, ci = 448 J/kg. ºC .)

Solution: Q cold = −Q hot

(mc∆T )w ater = − (mc∆T )iron 20.0 kg (4 186 J kg ⋅ °C )(T f

)

(

− 25.0°C = − (1.50 kg )(448 J kg ⋅ °C ) T f − 600°C

)

T f = 29.6°C

10.

How much heat is given up when 20.0 g of steam at 100.0 ºC is condensed and cooled to 20.0 ºC? (Given, c = 1.00 cal/g. ºC, L = 540 cal/g)

Solution: Heat change = condensation heat change + heat change of water during cooling = mLv + mc ∆T = [(20g) (-540 cal/g)] + [(1.00 cal/g. ºC) (20g) (20 ºC – 100 ºC) = - 12 400 cal = -12.4 kcal.

11.

How much energy is required to change a 40.0-g ice cube from ice at -10.0 ºC to steam at 110 ºC?

(Lf= 3.33 x 105 J/kg, ci= 2090 J/kg. ºC, Lv= 2.26 x 106 J/kg, cw= 4186 J/kg. ºC and cs= 2010 J/kg. ºC) Solution: The heat needed is the sum of the following terms: Q need ed = (heat to reach m elting point ) + (heat to m elt )

+ (heat to reach boiling point ) + (heat to vaporize ) + (heat to reach 110°C )

Thus, we have

(

Q need ed = 0.040 0 kg (2 090 J kg ⋅ °C )(10.0°C ) + 3.33 × 10 5 J kg

(

)

)

+ (4 186 J kg ⋅ °C )(100°C ) + 2.26 × 10 J kg + (2 010 J kg ⋅ °C )(10.0°C ) 6

Q need ed = 1.22 × 10 J 5

12.

The PETRONAS CARIGALI pipeline is 1 300 km long, reaching from Duyung Bay to the port of Miri. It experiences temperatures from –73°C to +35°C. How much does the steel pipeline expand because of the difference in temperature? (The coefficient of linear expansion, α= 11 x 10-6 (°C)-1)

Solution: ∆L = α Li ∆T = 11 × 10−6 (°C )

−1

13.

(1 300 km )[35°C − (−73°C )] =

1.54 km

The active element of a certain laser is made of glass rod 30.0cm long by 1.50cm in diameter. If the temperature of the rod increases by 65.0°C, what is the increase in; (a) its length (b) its volume? Assume that the average coefficient of linear expansion is 9.00 x 10-6 (°C)-1.

Solution: (a)

(b)

14.

∆L = α Li ∆T = 9.00 × 10−6 °C −1 (30.0 cm )(65.0°C ) = 0.176 mm

 30.0 (π )(1.50 )2  ∆V = 3α V i ∆T = 3 9.00 × 10−6 °C −1  cm 3  (65.0°C ) = 0.093 0 cm 3 4  

(

)

A Flat bimetallic strip consists of aluminum riveted to a strip of iron. When heated, the strip will bend. Which metal will be on the outside of the curve? Why? (αAl= 25 x 10-6 (°C)-1, αiron= 12 x 10-6 (°C)-1

. Solution: When heated, the aluminum expands more than the iron, because the expansion coefficient of a aluminum is larger than that of iron. Thus the aluminum will be on the outside of the curve.

15.

A cylinder of diameter 1.000 00 cm at 30.0 ºC is to be slid into a hole in a steel plate. The hole has a diameter of 0.999 70 cm at 30.0 ºC. To what temperature must the plate be heated? (Given, αsteel = 1.1 x 10-5 ºC-1).

Solution: The plate will expand in the same way whether or not there is a hole in it. Hence, the hole expands in the same way a circle of steel filling it would expand. So, the diameter change, ∆L = ( 1.000 00 – 0. 999 70 )cm = 0.000 30 cm. Using, ∆L = αLo ∆T Then, ∆T = 0.000 30 cm / [(1.1 x 10-5 ºC-1)( 0.999 70 cm)] = 27.0 ºC The temperature of the plate must be 30.0 + 27.0 = 57.0 ºC