Home Work Solutions PHYS111.02 SFSU Eradat [WA16 CHAPTERS 16] 1 Chapter 16 5,7,11,17,18,25,28,29 5. Picture the Problem: The temperature of the surface of the Sun is given in Kelvin and needs to be converted to Celsius and Fahrenheit. Strategy: Use equation 16-3 to convert the temperature to Celsius. Then insert the result into equation 16-1 to
calculate the temperature in Fahrenheit.
Solution: 1. (a) Convert from Kelvin to Celsius: TC = T − 273.15 K = 6000 K − 273.15 K = 5.7 × 103 °C
TF =
2. (b) Insert TC into equation 16-­‐1: (
)
9
5727 °C + 32 °F = 1.0 × 104 °F 5
Insight: The surface of the Sun is greater than 10,000°F! 7. Picture the Problem: The plot shows the pressure of the gas as a function of temperature. Note that at T2 = 105°C the pressure is P2 = 93.5 kPa, and at −273.15°C the pressure extrapolates linearly to zero. We also indicate the points corresponding to temperature of T1 = 50°C, and the point corresponding to a pressure of P3 = 115 kPa. Strategy: We assume that the pressure lies on a straight line. Using the known pressures at T2 = 105°C and at −273.15°C, calculate the rate at which pressure increases as a function of temperature. This rate can then be used to calculate the pressure at any temperature, and the temperature at a specified pressure. Solution: 1. Calculate the rate that pressure increases with temperature: rate =
2. (a) Use the rate to find the pressure at 50°C: P1 = 0.2473 kPa/C° ⎡⎣50°C − −273.15°C ⎤⎦ = 79.9 kPa 3. (b) Solve the rate equation for the temperature as a function of pressure: P3 = rate T3 − T0
93.5 kPa
= 0.2473 kPa/C° 105°C − −273.15°C
(
)
(
)
( )(
T3 =
P3
( rate )
+ T0 =
(
)
)
115 kPa
+ −273.15°C = 192 °C
0.2473 kPa/C°
Insight: As expected the pressure at 50°C is less than the pressure at 105°C, and the temperature at 115 kPa is greater than the temperature at 93.5 kPa. 11. Picture the Problem: Our sketch shows the pressure in the gas thermometer as a function of the temperature. Note that at T = 100°C the pressure is 227 mmHg and the pressure extrapolates to zero at the temperature −273.15°C. Assuming ideal behavior we want to find the temperature associated with a pressure of 162 mmHg. 2 Strategy: We assume that the pressure lies on a straight line. Using the known pressures at 100°C and −273.15°C, calculate the rate at which pressure increases as a function of temperature. Calculate the temperature at 162 mmHg using this rate. Solution: 1. Divide the change in pressure by the change in temperature to obtain the rate: rate =
227 mmHg − 0
= 0.60833 mmHg/C° 100 °C − −273.15°C
2. Solve the rate equation for the temperature: rate =
P − P0
T − T0
3. Insert given data: T = 100°C +
(
)
⇒ T = T0 +
P − P0 rate
162 mmHg − 227 mmHg
= − 6.85°C 0.60833 mmHg/C°
Insight: The gas thermometer is first placed in the boiling water to calibrate it, because water boils at a known temperature. Once the thermometer is calibrated, the pressure variation can accurately give temperatures over a wide range. 17. Picture the Problem: The Akashi Kaikyo Bridge in Japan is made of steel. When steel is heated it expands and when it is cooled it contracts. Strategy: In this problem we wish to find the change in length of the bridge between a cold winter day and a warm summer day. Use equation 16-­‐4 to determine the change in length. The coefficient of linear expansion for steel is given in Table 16-­‐1. Solution: Insert the given values into equation 16-­‐4: ΔL = α L0 ΔT
(
)
= ⎡⎣1.2 × 10−5 (C°)−1 ⎤⎦ 3910 m ⎡⎣30.0 °C − (−5.00 °C) ⎤⎦ = 1.6 m
Insight: This change in length is about the height of a person. If there were no expansion joints in the bridge this increase in length would be sufficient to buckle the bridge. 18. Picture the Problem: An aluminum plate has a hole cut in its center. The plate expands as it is heated. Strategy: We want to find the size of the hole after the temperature has increased to 199.0°C. The hole will expand at the same rate as the aluminum. Since the diameter of the hole is a unit of length, use equation 16-­‐4 to calculate the diameter as a function of the increase in temperature. The coefficient of linear expansion is given in Table 16-­‐1. Solution: 1. (a) Solve equation 16-­‐4 for the final diameter: Δd = d ′ − d = α dΔT
2. Insert the given data: d ′ = 1.178 cm ⎡⎣1+ 24 × 10 –6 K −1 199.0 °C − 23.00 °C ⎤⎦ = 1.183 cm
3. (b) Solve equation 16-­‐4 for the change in temperature: Δd = d ′ − d = α dΔT
d′ − d
ΔT = T − T0 =
αd
(
d ′ = d + α dΔT = d 1 + α ΔT
(
) )(
)
Home Work Solutions PHYS111.02 SFSU Eradat [WA16 CHAPTERS 16] 3 4. Solve for the final temperature: d′ − d
1.176 cm − 1.178 cm
= 23.00 °C +
= − 48°C αd
24 × 10 –6 K −1 1.178 cm
T = T0 +
(
)(
)
Insight: Since the final diameter (1.176 cm) is smaller than the diameter at 23°C we would expect that the final temperature would be below 23°C. The calculations show that this is the case. 25. Picture the Problem: The volume of an aluminum saucepan expands as the temperature of the pan increases. Water, which initially fills the saucepan to the brim, also increases in temperature and expands. If the water expands more than the saucepan, the water will spill over the top. If the saucepan expands more than the water, the water level will drop below the brim of the pan. Strategy: Use equation 16-­‐6 to calculate the change in volumes of the saucepan and of the water. Subtract the change in volume of the saucepan from the change in volume of the water to determine the volume of water that overflows the saucepan. The coefficient of volume expansion for water is given in Table 16-­‐1. The coefficient of volume of expansion of aluminum is three times its coefficient of linear expansion, which is also given in Table 16-­‐1. Solution: 1. (a) Because water has a larger coefficient of volumetric expansion, its volume will increase more than the volume of the aluminum sauce pan. Therefore, water will overflow from the pan. 2. (b) Calculate the initial volumes of the saucepan and water: ⎛ 23 cm ⎞
V0 = π r h = π ⎜
⎝ 2 ⎟⎠
3. Use equation 16-­‐6 to write the changes in volume: ΔVw = β wV0 ΔT
4. Subtract the change on volume of the pan from the water to calculate the volume of water spilled: Vspill = ΔVw − ΔVAl = ( β w − 3α Al )V0 ΔT
2
ΔVAl = 3α AlV0 ΔT
2
(6.0 cm ) = 2493c m
3
(
)
( ) ( 2493 cm ) ⎡⎣(88 − 19) °C ⎤⎦
= 21− 3 × 2.4 × 10−5 C°
−1
3
Vspill = 24 cm 3
Insight: This is the same principle that enables a mercury thermometer to work. The mercury expands faster than the surrounding glass, causing the mercury column to rise. 28. Picture the Problem: A person is lifting weights during a workout. The person does work against gravity each time the weight is lifted. Strategy: Calculate the amount of work done each time the weight is lifted and convert the results to calories. Divide the total work done by the work per lift to calculate the number of lifts necessary to expend the specified amount of calories. Solution: 1. Multiply force by distance W = F Δy = 12 lb 1.3 ft = 15.6 ft-lb to calculate work done in each repetition: ⎛ 1 m ⎞ ⎛ 4.448 N ⎞ ⎛ 1 Cal ⎞
2. Convert from ft-­‐lbs to Calories: W = 15.6 lb ⋅ ft ⎜
= 5.05 × 10−3 Cal
⎝ 3.281 ft ⎟⎠ ⎜⎝ lb ⎟⎠ ⎜⎝ 4186 N ⋅ m ⎟⎠
(
(
)(
)
)
3. Divide the total energy by the energy per repetition: Insight: Note that there are about 150 Calories in one-­‐half of a standard size Snicker’s Bar®. In order to do 30,000 repetitions to “work off” this half a candy bar, at the rate of 1 repetition every 2 seconds, it will take you almost 17 h! repetitions =
150 Cal
= 3.0 × 104 5.05 × 10−3 Cal
4 29. Picture the Problem: The figure shows Joule’s apparatus. As the weights fall at constant speed, the work done is converted into thermal energy, raising the temperature of the water. Strategy: We need to calculate the increase in temperature for a given amount of work. Calculate the work done by gravity on the weights as they fall through the height h. Set this work equal to the increase in thermal energy of the water. Multiply the result by 1.0 C° 6200 J to (
)
determine the increase in temperature. )(
) (0.48 m ) = 8.95 J Solution: 1. (a) Calculate the work done by gravity on the two weights: W = mgh = ⎡⎣ 2 0.95 kg ⎤⎦ 9.81 m/s
2. Multiply the work by the change in temperature per work: ⎛ 1.0 C° ⎞
ΔTC = ⎜
8.95 J = 1.4 × 10−3 C° ⎟
6200
J
⎝
⎠
3. (b) Fahrenheit degrees are smaller than Celsius degrees, so the temperature rise in Fahrenheit degrees would be greater than the result in part (a). 4. (c) Convert to Fahrenheit degrees: (
(
ΔTF =
(
2
)
)
9 F°
1.44 × 10−3 C° = 2.6 × 10−3 F° 5 C°
Insight: This change in temperature was not measurable. Joule had to repeatedly raise and lower the weights to achieve a measurable temperature difference.