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Convert the following temperatures from \(\mathrm{K}\) to \({ }^{\circ} \mathrm{C}\) (a) \(293.15 \mathrm{~K}\), (b) \(233.15 \mathrm{~K}\), (c) \(533.15 \mathrm{~K}\), (d) \(255.4 \mathrm{~K}\), (e) \(373.15 \mathrm{~K}\), (f) \(0 \mathrm{~K}\).

Short Answer

Expert verified
293.15 K = 20 °C, 233.15 K = -40 °C, 533.15 K = 260 °C, 255.4 K = -17.75 °C, 373.15 K = 100 °C, 0 K = -273.15 °C.

Step by step solution

01

- Understand the formula

To convert temperatures from Kelvin (K) to Celsius (°C), use the formula: \[ T(°C) = T(K) - 273.15 \]
02

- Convert 293.15 K to °C

Using the formula, subtract 273.15 from 293.15: \[ T(°C) = 293.15 - 273.15 = 20 \]
03

- Convert 233.15 K to °C

Using the formula, subtract 273.15 from 233.15: \[ T(°C) = 233.15 - 273.15 = -40 \]
04

- Convert 533.15 K to °C

Using the formula, subtract 273.15 from 533.15: \[ T(°C) = 533.15 - 273.15 = 260 \]
05

- Convert 255.4 K to °C

Using the formula, subtract 273.15 from 255.4: \[ T(°C) = 255.4 - 273.15 = -17.75 \]
06

- Convert 373.15 K to °C

Using the formula, subtract 273.15 from 373.15: \[ T(°C) = 373.15 - 273.15 = 100 \]
07

- Convert 0 K to °C

Using the formula, subtract 273.15 from 0: \[ T(°C) = 0 - 273.15 = -273.15 \]

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Key Concepts

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

Temperature Conversion
Temperature conversion is a crucial concept in thermodynamics and daily life. To effectively convert temperatures between different scales, like from Kelvin to Celsius, understanding the relationship between these scales is essential. The Kelvin to Celsius conversion is quite simple.
Use the formula: \( T(°C) = T(K) - 273.15 \).
For example: If you have a temperature of 293.15 K and want to convert it to Celsius, you subtract 273.15 from it, yielding 20 °C.
This helps us understand the temperature in a context more familiar to many people, the Celsius scale, when given in Kelvin.
Thermodynamics Education
Thermodynamics is a branch of physics that deals with heat and temperature and their relation to energy and work. It's foundational for various scientific disciplines, including chemistry, engineering, and environmental science.
Understanding temperature scales and conversions is vital in thermodynamics education. Practicing problems like converting Kelvin to Celsius strengthens comprehension of thermal energy and its behavior in different systems.
Students often encounter these concepts in labs and real-world applications, making these conversions practical and necessary for deeper understanding.
Kelvin Scale
The Kelvin scale is an absolute temperature scale used worldwide in scientific contexts. It starts from absolute zero, the coldest possible temperature, where molecular motion ceases.
Absolute zero is 0 K, which equates to -273.15 °C. Unlike Celsius and Fahrenheit, Kelvin doesn't use the term 'degrees.'
Key points about the Kelvin scale:
  • It's an SI unit, absolute scale.
  • 0 K is absolute zero.
  • Water freezes at 273.15 K, boils at 373.15 K.
This scale is crucial for scientific measurements because it provides a consistent and absolute measure of thermal energy.
Celsius Scale
The Celsius scale, or centigrade scale, is widely used in most of the world for everyday temperature measurements. It is based on the freezing and boiling points of water at standard atmospheric pressure, where 0 °C is the freezing point and 100 °C is the boiling point.
Key points about the Celsius scale:
  • It's convenient for daily use.
  • Widely adopted outside the USA.
  • Easily related to common experiences (e.g., weather).
Understanding Celsius helps in recognizing everyday situations and performing conversions with other temperature scales like Kelvin or Fahrenheit.
SI Units
The International System of Units (SI) is the standard for measurements worldwide, ensuring consistency and clarity in scientific communication. Temperature, in SI units, is measured in Kelvin (K).
Key points about SI units:
  • Globally recognized standards.
  • Ensures uniformity in scientific data.
  • Promotes efficient communication in international collaborations.
Learning SI units like Kelvin alongside Celsius enables a broader understanding of temperature measurements, crucial for scientific endeavors and global applications.

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Most popular questions from this chapter

In a steam power station working on the ideal Rankine cycle with regeneration, steam enters the turbine at 150 bar, \(600^{\circ} \mathrm{C}\). One open feedwater heater is used in the plant. Some steam from the turbine enters the open feedwater heater at a pressure of 12 bar. The pressure in the condenser is \(0.1\) bar. Determine the rate of exergy input to the working fluid passing through the steam generator, in \(\mathrm{kJ} /\) \(\mathrm{kg}\) of steam entering the turbine. Let \(T_{0}=288 \mathrm{~K}\) and \(p_{0}=1\) bar. Also, determine the rate of exergy destruction in the open feedwater heater, in \(\mathrm{kJ} / \mathrm{kg}\) of steam entering the turbine.

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Based on the macroscopic view, a quantity of air at \(100 \mathrm{kPa}\), \(20^{\circ} \mathrm{C}\) is in equilibrium. Yet the atoms and molecules of the air are in constant motion. How do you reconcile this apparent contradiction?

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