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Chapter 1: Problem 24

The temperature of a body on Kelvin scale is found to be \(X K\). When it is measured by a Fahrenheit thermometer, it is found to be \(X F\). Then \(X\) is (a) \(301.25\) (b) \(574.25\) (c) 313 (d) 40

Short Answer

Expert verified
The value of \(X\) such that the temperature is the same in both Kelvin and Fahrenheit scales is \(574.25\). Therefore, the correct answer is option (b) \(574.25\).

Step by step solution

01

Write down the Equation

Start with the equation for converting Kelvin to Fahrenheit: \(F = \frac{9}{5}(K-273) + 32\) Since the problem tells us that the temperature is the same in both scales, express \(K\) and \(F\) as \(X\). Substituting in the given values, the equation becomes: \(X = \frac{9}{5}(X-273) + 32\)
02

Solve the Equation

Now, the task is to solve this equation for \(X\). This is a linear equation in one variable, and it can be solved using basic algebraic operations. First, subtract 32 from both sides of the equation to get: \(X - 32 = \frac{9}{5}(X - 273)\). Multiply through by 5 to get: \(5X - 160 = 9(X - 273)\). Distribute the 9 on the right side to get \(5X - 160 = 9X - 2457\). Move the terms involving \(X\) to one side of the equation and the constants to the other to get: \(4X = 2297\). Finally, divide by 4 on both sides to solve for \(X\): \(X = \frac{2297}{4} = 574.25\)
03

Compare with the Options

The calculated value of \(X\) is 574.25. Comparing this with the provided options, it is seen that \(X = 574.25\) is indeed one of the options. Therefore, the correct answer is option (b) \(574.25\).

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Key Concepts

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

Kelvin scale
The Kelvin scale is a fundamental unit of temperature measurement used primarily in scientific settings. It is an absolute temperature scale, meaning it starts at absolute zero, the theoretical point where particles have minimal thermal motion. Unlike Celsius and Fahrenheit, there are no negative temperatures on the Kelvin scale, making it highly useful in scientific calculations and experiments.
- The Kelvin scale measures temperatures starting from absolute zero, represented as 0 K.
- Each increment on the Kelvin scale is the same size as a degree Celsius, which makes conversions between these two scales straightforward.
- Importantly, Kelvin does not use the degree symbol (掳), only 'K' is used following numerals. Since Kelvin is an absolute scale, it is crucial in thermodynamics, where laws often require measurements in this unit for accurate calculations. Scientists prefer using Kelvin to ensure consistency in temperature-based equations like those used in predicting gas behavior or thermodynamic processes.
Fahrenheit scale
The Fahrenheit scale, known for its more fine-tuned gradations in daily weather forecasts and household settings, is primarily used in the United States and some Caribbean countries. Unlike the other temperature scales, the Fahrenheit scale has its origins in the early 18th century.
- Daniel Gabriel Fahrenheit, a Polish-German physicist, invented this scale in 1724.
- In Fahrenheit, water freezes at 32掳F and boils at 212掳F under standard atmospheric pressure, offering a wider scale between the freezing and boiling points of water compared to Celsius. The choice of the Fahrenheit scale highlights a different method of expressing temperature changes than other scales, making it intuitive for those accustomed to its usage. Although less common in scientific circles than the Kelvin or Celsius scales, Fahrenheit remains relevant in day-to-day temperature recordings in places where it is adopted.
Temperature conversion
Temperature conversion is essential for comparing temperature readings from different scales and is frequently needed in science, engineering, and daily life. The conversion from Kelvin to Fahrenheit, in particular, is less direct compared to Celsius conversions due to the different zero points and unit sizes in these scales.
To convert a temperature from Kelvin (\( K \)) to Fahrenheit (\( F \)), the formula is:
\[ F = \frac{9}{5}(K - 273) + 32 \]
Here's a breakdown to understand this formula better:- The \( K - 273 \) accounts for the conversion from Kelvin to Celsius since 273 is the difference in zero points between Kelvin and Celsius.- Multiplying by \( \frac{9}{5} \) scales the Celsius degree to a Fahrenheit degree.- Finally, adding 32 adjusts for the difference in the baseline temperature of the Fahrenheit scale.Understanding these conversion principles is crucial for working across different disciplines where temperature needs to be universally understood, allowing for seamless integration of findings and applications.

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Most popular questions from this chapter

In two systems of relations among velocity, acceleration and force are respectively \(v_{2}=\frac{\alpha^{2}}{\beta} v_{1}\), \(a_{2}=\alpha \beta a_{1}\) and \(F_{2}=\frac{F_{1}}{\alpha \beta}\). If \(\alpha\) and \(\beta\) are constants then relations among mass, length and time in two systems are (a) \(M_{2}=\frac{\alpha}{\beta} M_{1}, L_{2}=\frac{\alpha^{2}}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha^{3} T_{1}}{\beta}\) (b) \(M_{2}=\frac{1}{\alpha^{2} \beta^{2}} M_{1}, L_{2}=\frac{\alpha^{3}}{\beta^{3}} L_{1}, T_{2}=T_{1} \frac{\alpha}{\beta^{2}}\) (c) \(M_{2}=\frac{\alpha^{3}}{\beta^{3}} M_{1}, L_{2}=\frac{\alpha^{2}}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha}{\beta} T_{1}\) (d) \(M_{2}=\frac{\alpha^{2}}{\beta^{2}} M_{1}, L_{2}=\frac{\alpha}{\beta^{2}} L_{1}, T_{2}=\frac{\alpha^{3}}{\beta^{3}} T_{1}\)

If the present units of length, time and mass \((m, s, k g)\) are changed to \(100 m, 100 s\), and \(\frac{1}{10} k g\) then (a) The new unit of velocity is increased 10 times (b) The new unit of force is decreased \(\frac{1}{1000}\) times (c) The new unit of energy is increased 10 times (d) The new unit of pressure is increased 1000 times

A physical quantity is measured and its value is found to be \(n u\) where \(n=\) numerical value and \(u=\) unit. Then which of the following relations is true (a) \(n \propto u^{2}\) (b) \(n \propto u\) (c) \(n \propto \sqrt{u}\) (d) \(n \propto \frac{1}{u}\)

A physical quantity \(P\) is given by \(P=\frac{A^{3} B^{\frac{1}{2}}}{C^{-4} D^{\frac{3}{2}}}\). The quantity which brings in the maximum percentage error in \(P\) is (a) \(A\) (b) \(B\) (c) \(C\) (d) \(D\)

The potential energy of a particle varies with distance \(x\) from a fixed origin as \(U=\frac{A \sqrt{x}}{x^{2}+B}\), where \(A\) and \(B\) are dimensional constants then dimensional formula for \(A B\) is (a) \(M L^{7 / 2} T^{-2}\) (b) \(M L^{11 / 2} T^{-2}\) (c) \(M^{2} L^{9 / 2} T^{-2}\) (d) \(M L^{13 / 2} T^{-3}\)
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