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Chapter 3: Q10E (page 574)

Like the Kelvin scale, the Rankinescaleis an absolute temperature scale: Absolute zero is zero degrees Rankine (0ºR). However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

Short Answer

Expert verified

The numerical value of the triple-point temperature of water on the Rankine scale is491.69°R.

Step by step solution

01

Concept of Kelvin Scale or Absolute Temperature Scale.

Kelvin Scale is specified using only one temperature as a reference point on the scale. The temperature chosen is the triple point of water and the scale is called Absolute Temperature Scale with T=0 K at -273.15ºC.The figure below shows the relationships between the Kelvin, Fahrenheit and Celsius scales.

02

 Determination of the numerical value of the triple-point temperature of water on the Rankine scale.

On the Kelvin scale,∆Tis same in Kand °Cscale and the triple point is 273.16 K

So that,

1K=1°Cand1°C=95F°,so1K=95R°

Thus, the triple point in Rankine scale is,

Ttriple=95273.16°R=491.69°R

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Most popular questions from this chapter

You hold an inflated balloon over a hot-air vent in your house and watch it slowly expand. You then remove it and let it cool back to room temperature. During the expansion, which was larger: the heat added to the balloon or the work done by the air inside it? Explain. (Assume that air is an ideal gas.) Once the balloon has returned to room temperature, how does the net heat gained or lost by the air inside it compare to the net work done on or by the surrounding air?

A liquid is irregularly stirred in a well-insulated container and thereby undergoes a rise in temperature. Regard the liquid as the system. Has heat been transferred? How can you tell? Has work been done? How can you tell? Why is it important that the stirring is irregular? What is the sign of ∆U? How can you tell?

Question: If the air temperature is the same as the temperature of your skin (about 30°C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70-kg person’s body produces energy at a rate of about 500W due to metabolism, 80% of which is converted to heat. (a) How many kilograms of water must the person’s body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is 2.42×106J/kg. (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a litre of water is 1.0 kg.

Section 18.1 states that ordinarily, pressure, volume, and temperature cannot change individually without one affecting the others. Yet when a liquid evaporates, its volume changes, even though its pressure and temperature are constant. Is this inconsistent? Why or why not?

-18.22 (a) If you apply the same amount of heat to 1.00 mol of an ideal monatomic gas and 1.00 mol of an ideal diatomic gas, which one (if any) will increase more in temperature? (b) Physically, why do diatomic gases have a greater molar heat capacity than mono-atomic gases?

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