(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,

we subtract the second equation from the first to obtain

First, we take TL = 373.125 K (the boiling point of water) and TR = 273.16 K (the triple point of water). Then, pLpR = 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 pLpR = 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

(a) The difference is pN pH = 0.056 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 . For x = –71°C, this gives y = –96°F.

(b) The relationship between y and x may be inverted to yield . 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 . If we require y = 2x, then we have

which yields y = 2x = 320°F.

(b) In this case, we require and find

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 = 40.00/500.0 = 8.000  10–2 and b = –60.00. With these values, we find x for y = 50.00:

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 = 3VT, where V is the original volume and is the coefficient of linear expansion. See Eq. 18-11. Since V = (4/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 for the alloy is

Thus, from 100°C to 0°C we have

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

giving T = 48 °C. Thus, Tx = (20°C + 48 °C )= 68°C.

11. 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 = 3.

14. (a) We use = m/V and

The percent change in density is

(b) Since = L/(LT ) = (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, a 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 = 3a VcT. See Eq. 18-11. If is the coefficient of volume expansion for glycerin then the change in the volume of glycerin is Vg = VcT. 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 20.11 cm mark is

Thus, the actual change in length for the rod is

L = (20.11 cm – 20.05 cm) + 0.055 cm = 0.115 cm.

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

17. After the change in temperature the diameter of the steel rod is Ds = Ds0 + sDs0T and the diameter of the brass ring is Db = Db0 + bDb0T, where Ds0 and Db0 are the original diameters, s and b 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

(Parte 1 de 8)