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Chapter 3: Problem 63

Convert the temperatures in Parts (a) and (b) and temperature intervals in Parts (c) and (d): (a) \(T=85^{\circ} \mathrm{F}\) to \(^{\circ} \mathrm{R},^{\circ} \mathrm{C}, \mathrm{K}\) (b) \(T=-10^{\circ} \mathrm{C}\) to \(\mathrm{K},^{\circ} \mathrm{F},^{\circ} \mathrm{R}\) (c) \(\Delta T=85^{\circ} \mathrm{C}\) to \(\mathrm{K},^{\circ} \mathrm{F},^{\circ} \mathrm{R}\) (d) \(\Delta T=150^{\circ} \mathrm{R}\) to \(^{\circ} \mathrm{F},^{\circ} \mathrm{C}, \mathrm{K}\)

Short Answer

Expert verified
The temperatures are converted as follows: (a) \(85^{\circ} \mathrm{F}\) is equal to \(544.67^{\circ} \mathrm{R}, 29.44^{\circ} \mathrm{C}, 302.59 \mathrm{K}\). (b) \(-10^{\circ} \mathrm{C}\) is equal to \(263.15 \mathrm{K}, 473.67^{\circ} \mathrm{R}, 14^{\circ} \mathrm{F}\). The temperature intervals transformed are: (c) \(\Delta T = 85^{\circ} \mathrm{C}\) is equal to \(85 \mathrm{K}, 153^{\circ} \mathrm{F}, 153^{\circ} \mathrm{R}\). (d) \(\Delta T = 150^{\circ} \mathrm{R}\) is equivalent to \(150^{\circ} \mathrm{F}, 83.33^{\circ} \mathrm{C}, 83.33 \mathrm{K}\).

Step by step solution

01

Conversion of Fahrenheit to other units

Starting with \(T=85^{\circ} \mathrm{F}\). To convert this to Rankine, use the formula \(T_R = T_F + 459.67\). To Celsius, the formula is \(T_C = (T_F - 32) \times 5/9\). To Kelvin, first convert to Celsius and then add 273.15.
02

Applying formulas to Part (a)

Applying these formulas, the converted temperatures are \(T_R = 85^{\circ} \mathrm{F} + 459.67 = 544.67^{\circ} \mathrm{R}\), \(T_C = (85^{\circ} \mathrm{F} - 32) \times 5/9 = 29.44^{\circ} \mathrm{C}\), and \(T_K = 29.44^{\circ} \mathrm{C} + 273.15 = 302.59 \mathrm{K}\).
03

Conversion of Celsius to other units

For \(T=-10^{\circ} \mathrm{C}\), to convert this temperature to Kelvin, Rankine, and Fahrenheit. The formulas are \(T_K = T_C + 273.15\), \(T_R = (T_C + 273.15) \times 9/5\), and \(T_F = T_C \times 9/5 + 32\), respectively.
04

Applying formulas to Part (b)

Using these formulas, \(T_K =-10^{\circ} \mathrm{C}+273.15 = 263.15 \mathrm{K}\), \(T_R = (263.15) \times 9/5 = 473.67^{\circ} \mathrm{R}\), and \(T_F = -10^{\circ} \mathrm{C} \times 9/5 + 32 = 14^{\circ} \mathrm{F}\).
05

Conversion of Celsius interval to other units

For \(\Delta T=85^{\circ} \mathrm{C}\), the interval in Kelvin is the same as in Celsius. To convert to Rankine and Fahrenheit, the same conversion factors as for temperature conversion are used: \(9/5\) and \(5/9\), respectively.
06

Applying formulas to Part (c)

Applying this, \(\Delta T = 85 \mathrm{K}\), \(\Delta T = 85 \mathrm{C} \times 9/5 = 153^{\circ} \mathrm{F}\), and \(\Delta T = 85 \mathrm{C} \times 9/5 = 153^{\circ} \mathrm{R}\). Note the interval stays the same in Kelvin and Celsius.
07

Conversion of Rankine interval to other units

The problem to solve in part (d) is to transform the \(\Delta T=150^{\circ} \mathrm{R}\) interval. For the intervals in Fahrenheit, Celsius and Kelvin, the conversion factors are \(5/9\) and \(9/5\). Note that for the Rankine and Fahrenheit scales the intervals are the same.
08

Applying formulas to Part (d)

After performing the conversions, \(\Delta T = 150^{\circ} \mathrm{F}\), \(\Delta T = 150^{\circ} \mathrm{R} \times 5/9 = 83.33^{\circ} \mathrm{C}\), and \(\Delta T = 83.33 \mathrm{K}\). Note the interval stays the same in Rankine and Fahrenheit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Unit conversion, especially in temperature measurements, is an essential skill in science and everyday life. When dealing with temperature conversion between different scales, understanding how to apply the correct formulas is vital. The key temperature scales we often convert between include Fahrenheit (掳F), Celsius (掳C), Kelvin (K), and Rankine (掳R).

For conversion:
  • Fahrenheit to Celsius: Subtract 32 from the Fahrenheit value and multiply by \(\frac{5}{9}\).
  • Celsius to Kelvin: Add 273.15 to the Celsius value.
  • Fahrenheit to Rankine: Add 459.67 to the Fahrenheit value.
  • Celsius to Fahrenheit: Multiply by \(\frac{9}{5}\) and add 32.
These conversions show how moving from one unit to another requires specific mathematical operations. Mastering these operations allows for accurate conversions in various scientific and practical applications.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. Temperature conversion is a fundamental part of thermodynamics because temperatures determine the direction of heat transfer.

Understanding thermodynamics involves recognizing how energy moves and changes form, especially in temperature exchanges. When converting temperature scales, it's part of understanding how energy behaves in different contexts, whether it's heating a substance or determining the energy output from chemical reactions.

In practice, thermodynamics often utilizes Kelvin because it starts at absolute zero, providing a clear basis for mathematical calculations, unlike Celsius or Fahrenheit, which are not absolute scales. This foundation is crucial in thermodynamic equations used to describe phenomena such as the laws of thermodynamics.
Temperature Scales
Temperature scales are systems for measuring temperature. They provide a standard for scientists and engineers to understand and communicate thermal conditions. Each scale has its use case, and understanding them is key to correctly interpreting temperature data.

  • Celsius: Used widely in most of the world, especially in science, it is based on the freezing and boiling points of water.
  • Fahrenheit: Mainly used in the United States, it provides a more granulated scale for temperature measurement.
  • Kelvin: Used in scientific contexts and is essential in thermodynamics, as it starts from absolute zero, the point theoretically devoid of all thermal energy.
  • Rankine: A scale used predominantly in certain engineering fields, which bears the same unit increments as Fahrenheit but starts from absolute zero.
Comprehending these scales is critical for unit conversion tasks, allowing you to convert temperatures appropriately and apply this knowledge in both everyday and scientific scenarios.

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Most popular questions from this chapter

An object of density \(\rho_{\mathrm{a}},\) volume \(V_{\mathrm{a}},\) and weight \(W_{\mathrm{a}}\) is thrown from a rowboat floating on the surface of a small pond and sinks to the bottom. The weight of the rowboat without the jettisoned object is \(W_{\mathrm{b}}\). Beforethe object was thrown out, the depth of the pond was \(h_{\mathrm{pl}}\), and the bottom of the boat was a distance \(h_{\mathrm{b} 1}\) above the pond bottom. After the object sinks, the values of these quantities are \(h_{\mathrm{p} 2}\) and \(h_{\mathrm{b} 2}\). The area of the pond is \(A_{\mathrm{p}}\); that of the boat is \(A_{b} . A_{b}\) may be assumed constant, so that the volume of water displaced by the boat is \(A_{\mathrm{b}}\left(h_{\mathrm{p}}-h_{\mathrm{b}}\right)\). (a) Derive an expression for the change in the pond depth \(\left(h_{\mathrm{p} 2}-h_{\mathrm{p} 1}\right) .\) Does the liquid level of the pond rise or fall, or is it indeterminate? (b) Derive an expression for the change in the height of the bottom of the boat above the bottom of the pond \(\left(h_{b 2}-h_{b 1}\right) .\) Does the boat rise or fall relative to the pond bottom, or is it indeterminate?

A housing development is served by a water tower with the water level maintained between 20 and 30 meters above the ground, depending on demand and water availability. Responding to a resident's complaint about the low flow rate of water at his kitchen sink, a representative of the developer measured the water pressure at the tap above the kitchen sink and at the junction between the water main (a pipe connected to the bottom of the water tower) and the feed pipe to the house. The junction is \(5 \mathrm{m}\) below the level of the kitchen tap. All water valves in the house were turned off. (a) If the water level in the tower was 25 m above tap level, what should be the gauge pressures (kPa) at the tap and junction? (b) Suppose the pressure measurement at the tap was lower than your estimate in Part (a), but the measurement at the junction was as predicted. State a possible explanation. (c) If pressure measurements corresponded to the predictions in Part (a), what else could be responsible for the low water flow to the sink?

Limestone (calcium carbonate) particles are stored in \(50-\mathrm{L}\) bags. The void fraction of the particulate matter is 0.30 (liter of void space per liter of total volume) and the specific gravity of solid calcium carbonate is 2.93. (a) Estimate the bulk density of the bag contents ( \(\mathbf{k g}\) CaCO \(_{3}\) /iter of total volume). (b) Estimate the weight ( \(W\) ) of the filled bags. State what you are neglecting in your estimate. (c) The contents of three bags are fed to a ball mill, a device something like a rotating clothes dryer containing steel balls. The tumbling action of the balls crushes the limestone particles and turns them into a powder. (See pp. \(21-64\) of Perry's Chemical Engineers' Handbook, 8 th ed.) The limestone coming out of the mill is put back into \(50-\mathrm{L}\) bags. Would the limestone (i) just fill three bags, (ii) fall short of filling three bags, or (iii) fill more than three bags? Briefly explain your answer.

A storage tank containing oil ( \(\mathrm{SG}=0.92\) ) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of oil it contains can be determined from the gauge pressure at the bottom. (a) A pressure gauge connected to the bottom of the tank was calibrated with the top of the tank open to the atmosphere. The calibration curve is a plot of height of oil, \(h(\mathrm{m}),\) versus \(P_{\text {geuge }}(\mathrm{kPa}) .\) Sketch the expected shape of this plot. What height of oil would lead to a gauge reading of \(68 \mathrm{kPa} ?\) What would be the mass (kg) of oil in the tank corresponding to this height? (b) An operator observes that the pressure gauge reading is 68 kPa and notes the corresponding liquid height from the calibration curve. What he did not know was that the absolute pressure above the liquid surface in the tank was \(115 \mathrm{kPa}\) when he read the gauge. What is the actual height of the oil? (Assume atmospheric pressure is \(101 \mathrm{kPa}\).)

In September 2014 the average price of gasoline in France was 1.54 euro/liter, and the exchange rate was \(\$ 1.29\) per euro ( \(\epsilon\) ). How much would you have paid, in dollars, for 50.0 kg of gasoline in France, assuming gasoline has a specific gravity of \(0.71 ?\) What would the same quantity of gasoline have cost in the United States at the prevailing average price of \(\$ 3.81 /\) gal?
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