# Resoluções Halliday 8ª Edição

(Parte 1 de 8)

Chapter 18

1. Let TL be the temperature and pL be the pressure in the left-hand thermometer.

Similarly, let TR be the temperature and pR be the pressure in the right-hand thermometer. According to the problem statement, the pressure is the same in the two thermometers when they are both at the triple point of water. We take this pressure to be p3. Writing Eq. 18-5 for each thermometer,

p p we subtract the second equation from the first to obtain

(273.16K) .LR L R p pT T

First, we take TL = 373.125 K (the boiling point of water) and TR = 273.16 K (the triple point of water). Then, pL – pR = 120 torr. We solve

for p3. The result is p3 = 328 torr. Now, we let TL = 273.16 K (the triple point of water) and TR be the unknown temperature. The pressure difference is pL – pR = 90.0 torr. Solving the equation

for the unknown temperature, we obtain TR = 348 K.

2. We take p3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 18-5 to compute the pressure:

The hydrogen thermometer gives 373.16 K for the boiling point of water and

CHAPTER 18

(a) The difference is pN − pH = 0.056 kPa 0.06 kPa≈.

(b) The pressure in the nitrogen thermometer is higher than the pressure in the hydrogen thermometer.

3. From Eq. 18-6, we see that the limiting value of the pressure ratio is the same as the absolute temperature ratio: (373.15 K)/(273.16 K) = 1.366.

4. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y. Then 9532yx=+. For x = –71°C, this gives y = –96°F.

(b) The relationship between y and x may be inverted to yield 59(32)xy=−. Thus, for y = 134 we find x ≈ 56.7 on the Celsius scale.

5. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y. Then 9532yx=+. If we require y = 2x, then we have

 9232 (5)(32)160C5

which yields y = 2x = 320°F. (b) In this case, we require 12yx= and find

 19(10)(32)32 24.6C

which yields y = x/2 = –12.3°F.

6. We assume scales X and Y are linearly related in the sense that reading x is related to reading y by a linear relationship y = mx + b. We determine the constants m and b by solving the simultaneous equations:

which yield the solutions m = 40.0/50.0 = 8.0 × 10–2 and b = –60.0. With these values, we find x for y = 50.0:

7. We assume scale X is a linear scale in the sense that if its reading is x then it is related to a reading y on the Kelvin scale by a linear relationship y = mx + b. We determine the constants m and b by solving the simultaneous equations:

which yield the solutions m = 100/(170 – 53.5) = 0.858 and b = 419. With these values, we find x for y = 340:

8. The change in length for the aluminum pole is

9. Since a volume is the product of three lengths, the change in volume due to a temperature change ∆T is given by ∆V = 3αV ∆T, where V is the original volume and α is the coefficient of linear expansion. See Eq. 18-1. Since V = (4pi/3)R3, where R is the original radius of the sphere, then

The value for the coefficient of linear expansion is found in Table 18-2. 10. (a) The coefficient of linear expansion α for the alloy is

The length at 0°C is therefore L′ = L + ∆L = (10.015 cm – 0.0188 cm) = 9.996 cm. (b) Let the temperature be Tx. Then from 20°C to Tx we have

CHAPTER 18 giving ∆T = 48 °C. Thus, Tx = (20°C + 48 °C )= 68°C.

1. The new diameter is

12. The increase in the surface area of the brass cube (which has six faces), which had side length is L at 20°, is

13. The volume at 30°C is given by

where we have used β = 3α. 14. (a) We use ρ = m/V and

(b) Since α = ∆L/(L∆T ) = (0.23 × 10–2) / (100°C – 0.0°C) = 23 × 10–6 /C°, the metal is aluminum (using Table 18-2).

15. If Vc is the original volume of the cup, αa is the coefficient of linear expansion of aluminum, and ∆T is the temperature increase, then the change in the volume of the cup is ∆Vc = 3αa Vc ∆T. See Eq. 18-1. If β is the coefficient of volume expansion for glycerin then the change in the volume of glycerin is ∆Vg = βVc ∆T. Note that the original volume of glycerin is the same as the original volume of the cup. The volume of glycerin that spills is

16. The change in length for the section of the steel ruler between its 20.05 cm mark and 20.1 cm mark is

Thus, the actual change in length for the rod is

∆L = (20.1 cm – 20.05 cm) + 0.055 cm = 0.15 cm.

The coefficient of thermal expansion for the material of which the rod is made is then

270C 20CL T

17. After the change in temperature the diameter of the steel rod is Ds = Ds0 + αsDs0 ∆T and the diameter of the brass ring is Db = Db0 + αbDb0 ∆T, where Ds0 and Db0 are the original diameters, αs and αb are the coefficients of linear expansion, and ∆T is the change in temperature. The rod just fits through the ring if Ds = Db. This means

Ds0 + αsDs0 ∆T = Db0 + αbDb0 ∆T. Therefore, s b

b b s s

The temperature is T = (25.00°C + 335.0 °C) = 360.0°C.

18. (a) Since A = piD2/4, we have the differential dA = 2(piD/4)dD. Dividing the latter relation by the former, we obtain dA/A = 2 dD/D. In terms of ∆'s, this reads

We can think of the factor of 2 as being due to the fact that area is a two-dimensional quantity. Therefore, the area increases by 2(0.18%) = 0.36%.

(b) Assuming that all dimensions are allowed to freely expand, then the thickness increases by 0.18%.

(c) The volume (a three-dimensional quantity) increases by 3(0.18%) = 0.54%.

CHAPTER 18

(d) The mass does not change. (e) The coefficient of linear expansion is

19. The initial volume V0 of the liquid is h0A0 where A0 is the initial cross-section area and h0 = 0.64 m. Its final volume is V = hA where h – h0 is what we wish to compute. Now, the area expands according to how the glass expands, which we analyze as follows:

(Parte 1 de 8)