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Chapter 10: Problem 15

Convert the following temperatures to absolute temperatures in kelvins: (a) \(0^{\circ} \mathrm{C}\) (b) \(100^{\circ} \mathrm{C}\) (c) \(20^{\circ} \mathrm{C},\) and (d) \(-35^{\circ} \mathrm{C}\).

Short Answer

Expert verified
0掳C = 273.15K, 100掳C = 373.15K, 20掳C = 293.15K, -35掳C = 238.15K.

Step by step solution

01

Identify the Conversion Formula

The formula to convert Celsius to Kelvin is: \( T(K) = T(^{\circ}C) + 273.15 \). This formula accounts for the offset between the Celsius scale and the Kelvin scale, where 0掳C is equivalent to 273.15K.
02

Convert 0掳C to Kelvin

Substitute 0 for \( T(^{\circ}C) \) in the formula: \[ T(K) = 0 + 273.15 = 273.15 \text{ K} \]. Thus, \(0^{\circ} \,C\) is 273.15 K.
03

Convert 100掳C to Kelvin

Substitute 100 for \( T(^{\circ}C) \) in the formula: \[ T(K) = 100 + 273.15 = 373.15 \text{ K} \]. Therefore, \(100^{\circ} \,C\) is 373.15 K.
04

Convert 20掳C to Kelvin

Substitute 20 for \( T(^{\circ}C) \) in the formula: \[ T(K) = 20 + 273.15 = 293.15 \text{ K} \]. Hence, \(20^{\circ} \,C\) is 293.15 K.
05

Convert -35掳C to Kelvin

Substitute -35 for \( T(^{\circ}C) \) in the formula: \[ T(K) = -35 + 273.15 = 238.15 \text{ K} \]. So, \(-35^{\circ} \,C\) is 238.15 K.

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Key Concepts

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

Celsius to Kelvin Conversion
Converting temperatures from Celsius to Kelvin is a straightforward process. The formula \( T(K) = T(^{\circ}C) + 273.15 \) enables us to make this transformation easily.
Notice that this formula simply adds 273.15 to the Celsius temperature. This number represents the difference between the two scales.
It is essential to recognize that the Celsius and Kelvin scales have the same magnitude of division; both have increments of 1 degree.
Thus, moving from one degree to the next in Kelvin is equivalent to moving one degree in Celsius.
This converter is widely used in scientific calculations to ensure uniformity.
Absolute Temperature
Absolute temperature, often measured in Kelvin, represents an absolute scale with no negative numbers. It starts at absolute zero.
Absolute zero (0 K) is the theoretical point where particles have minimal vibrational movement.
This scale is particularly useful in physics and engineering, where it is essential to compare thermal energies.
Using the Kelvin scale, we can express temperatures without the confusion of negative values, simplifying many calculations.
Kelvin Scale
The Kelvin scale is one of the most commonly used absolute temperature scales in science.
It begins at absolute zero, the point where molecular energy is at its minimum possible level.
This absolute scale is named after William Thomson, also known as Lord Kelvin, who introduced it.
One of its key features is its use in scientific research, particularly in fields like thermodynamics.
Scientists prefer it because it portrays true energy states without the complications of negative values.
Temperature Units
Temperature units like Celsius, Kelvin, and Fahrenheit help us describe how hot or cold something is.
While Celsius is commonly used in daily life and weather forecasts, Kelvin is preferred in scientific circles.
The distinction between these units lies in their starting points and scaling.
  • Celsius: Water freezes at 0掳C and boils at 100掳C.
  • Kelvin: Starts at absolute zero, with water freezing at 273.15 K and boiling at 373.15 K.
  • Fahrenheit: Mainly used in the United States, where water freezes at 32掳F and boils at 212掳F.
Each unit serves its purpose, making it vital to understand the context in which it is used.

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Most popular questions from this chapter

In the troposphere (the lowest part of the atmosphere), the temperature decreases rather uniformly with altitude at a so-called "lapse" rate of about \(6.5^{\circ} \mathrm{C} / \mathrm{km}\). What are the temperatures (a) near the top of the troposphere (which has an average thickness of \(11 \mathrm{~km}\) ) and (b) outside a commercial aircraft flying at a cruising altitude of \(34000 \mathrm{ft} ?\) (Assume that the ground temperature is normal room temperature.)

During open-heart surgery it is common to cool the patient's body down to slow body processes and gain an extra margin of safety. A drop of \(8.5^{\circ} \mathrm{C}\) is typical in these types of operations. If a patient's normal body temperature is \(98.2^{\circ} \mathrm{F}\), what is her final temperature in both Celsius and Fahrenheit?

An athlete has a large lung capacity, 7.0 L. Assuming air to be an ideal gas, how many molecules of air are in the athlete's lungs when the air temperature in the lungs is \(37^{\circ} \mathrm{C}\) under normal atmospheric pressure?

A steel-belted radial automobile tire is inflated to a gauge pressure of \(30.0 \mathrm{lb} / \mathrm{in}^{2}\) when the temperature is \(61^{\circ} \mathrm{F}\). Later in the day, the temperature rises to \(100^{\circ} \mathrm{F}\) Assuming the volume of the tire remains constant, what is the tire's pressure at the elevated temperature? [Hint: Remember that the ideal gas law uses absolute pressure.]

When lightning strikes, it can heat the air around it to more than \(30000 \mathrm{~K}\), five times the surface temperature of the Sun. (a) What is this temperature on the Fahrenheit and Celsius scales? (b) The temperature is sometimes reported to be \(30000^{\circ} \mathrm{C}\). Assuming that \(30000 \mathrm{~K}\) is correct, what is the percentage error of this Celsius value?
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