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Chapter 17: Problem 5

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 K. What is its temperature change in (a) F\(^\circ\) and (b) C\(^\circ\)?

Short Answer

Expert verified
10.0 K equals 10.0 C掳 and 18.0 F掳 temperature change.

Step by step solution

01

Understand the Problem

The problem provides a temperature change in Kelvin (10.0 K) and asks us to convert this change to Fahrenheit and Celsius.
02

Convert Kelvin Change to Celsius Change

Since a change of 1 Kelvin is equivalent to a change of 1 degree Celsius, a temperature change of 10.0 K is equal to a change of 10.0 掳C.
03

Convert Celsius Change to Fahrenheit Change

Use the conversion formula for temperature differences: \( \Delta T_{掳F} = \Delta T_{掳C} \times \frac{9}{5} \). Substitute \( \Delta T_{掳C} = 10.0 \) to get \( \Delta T_{掳F} = 10.0 \times \frac{9}{5} = 18.0 \).
04

Conclusion

The temperature change in Fahrenheit is 18.0 F掳 and in Celsius is 10.0 C掳.

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Key Concepts

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

Kelvin to Celsius
The Kelvin and Celsius scales are two fundamental ways to measure temperature. A remarkable feature of these two scales is how closely they relate to each other.
To convert a temperature difference from Kelvin to Celsius, there is actually nothing complex about it. This is because both scales increase at the same rate.
  • 1 degree Celsius is exactly equal to 1 Kelvin when it comes to a temperature change.
This means that if an object experiences a temperature change of 10.0 Kelvin, its change in degrees Celsius would also be 10.0 C掳.
This direct equivalence makes calculations between Kelvin and Celsius very straightforward. There's no need to add or subtract any constant鈥攋ust a simple 1-to-1 conversion!
Celsius to Fahrenheit
Sometimes, you need to express Celsius in Fahrenheit, especially in places where Fahrenheit is commonly used.
When dealing with temperature changes, the scale conversion requires a little formulaic adjustment.To convert a temperature difference from Celsius to Fahrenheit, use the formula:\[ \Delta T_{掳F} = \Delta T_{掳C} \times \frac{9}{5} \]This equation scales the Celsius change into Fahrenheit.
For example, if a temperature change is 10.0 掳C, then:
  • Plug into the formula: \(\Delta T_{掳F} = 10.0 \times \frac{9}{5} = 18.0\)
  • The temperature change is 18.0 Fahrenheit degrees (F掳).
This conversion is essential for countless scientific, engineering, and everyday practical applications, ensuring everyone can understand and communicate temperature changes efficiently.
Temperature Change
Temperature change refers to how much the temperature of an object or substance increases or decreases.
This is an essential concept in physics and everyday life because it affects how substances react or behave.Here are a few things to note about temperature changes:
  • Temperature change can occur in various scales such as Kelvin, Celsius, or Fahrenheit.
  • It is represented by the symbol \(\Delta T\), where \(\Delta\) signifies change.
  • It's crucial for understanding environmental shifts, chemical reactions, and physical transformations.
Whether you're chilling a beverage or studying thermodynamics, knowing how to properly calculate and express temperature changes helps you grasp the broader implications of heat transfer and energy conversion.

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Most popular questions from this chapter

If the air temperature is the same as the temperature of your skin (about 30\(^\circ\)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 500 W 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 \(2.42 \times 10{^6} J/kg\). (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 liter of water is 1.0 kg.)

On a cool (4.0\(^\circ\)C) Saturday morning, a pilot fills the fuel tanks of her Pitts S-2C (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again 4.0\(^\circ\)C, she checks the fuel level and finds only 103.4 L of gasoline in the aluminum tanks. She realizes that it was hot on Saturday afternoon and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in \(^\circ\)C) of the fuel and the tank on Saturday afternoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10{^-}{^4} K{^-}{^1}\). (b) To have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student鈥檚 body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to 44\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e = 1 for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10{^3}{^2} W\) and has surface temperature 11,000 K; (b) Procyon B (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10{^2}{^3} W\) and has surface temperature 10,000 K. (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon B is an example of a white dwarf star.)

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around 33\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0\(^\circ\)C (104\(^\circ\)F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.
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