/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 At what temperature is the Fahre... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius scale and (b) half that of the Celsius scale?

Short Answer

Expert verified
(a) 160°C, (b) -24.6°C approximately.

Step by step solution

01

Understand the Fahrenheit and Celsius relation

We know that the relationship between the Fahrenheit (F) and Celsius (C) scales is given by the formula: \[ F = \frac{9}{5}C + 32 \] This formula will help us determine at what temperature the conditions given in parts (a) and (b) are met.
02

Solving Part (a): Fahrenheit is Twice Celsius

For part (a), we want to find the temperature where the Fahrenheit reading is twice the Celsius reading, so we set: \[ F = 2C \] Substitute \( F \) in the formula: \[ 2C = \frac{9}{5}C + 32 \] Rearranging gives: \[ 2C - \frac{9}{5}C = 32 \] Finding a common denominator and solving for \( C \): \[ \frac{10}{5}C - \frac{9}{5}C = 32 \] \[ \frac{1}{5}C = 32 \] Multiply both sides by 5: \[ C = 160 \] When \( C = 160 \), \( F = 2 \times 160 = 320 \). Thus, the temperature at which the Fahrenheit reading is twice the Celsius reading is \( 160°C \).
03

Solving Part (b): Fahrenheit is Half Celsius

For part (b), we want to find the temperature where the Fahrenheit reading is half the Celsius reading, so we set: \[ F = \frac{1}{2}C \] Substitute \( F \) in the formula: \[ \frac{1}{2}C = \frac{9}{5}C + 32 \] Rearranging gives: \[ \frac{1}{2}C - \frac{9}{5}C = 32 \] Find a common denominator and solve for \( C \): \[ \frac{5}{10}C - \frac{18}{10}C = 32 \] \[ -\frac{13}{10}C = 32 \] Multiply both sides by \(-\frac{10}{13}\): \[ C = -\frac{320}{13} \] Convert to a decimal if needed: \( C \approx -24.6 \) Thus, the temperature at which the Fahrenheit reading is half the Celsius reading is approximately \(-24.6°C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Temperature Conversion
Temperature conversion between Fahrenheit and Celsius is a common necessity in science and daily life. The conversion formula is: \[ F = \frac{9}{5}C + 32 \] where \( F \) is the temperature in degrees Fahrenheit and \( C \) is the temperature in degrees Celsius. This formula emerges from the difference in how the two scales are set: Celsius sets 0 degrees at the freezing point of water and 100 degrees at the boiling point, while Fahrenheit sets these at 32 and 212 degrees, respectively. Converting from Celsius to Fahrenheit involves multiplying the Celsius value by \( \frac{9}{5} \) and then adding 32. This conversion is handy whether you're reading a scientific study that uses a different scale or checking the weather forecast when traveling abroad.
Temperature Scales
Temperature scales like Celsius, Fahrenheit, and others are systems of measuring temperature. Each scale has its own method of defining, measuring, and interpreting temperature differences.
  • Celsius Scale: Commonly used worldwide, this scale defines 0 degrees as the freezing point of water and 100 degrees as the boiling point under standard atmospheric pressure.
  • Fahrenheit Scale: Primarily used in the United States, this scale sets the freezing point of water at 32 degrees and the boiling point at 212 degrees.
Temperature scales are vital in scientific calculations and everyday life, providing a standard for weather reports, scientific studies, and more. Understanding multiple scales is important as it allows for accurate interpretation and conversion of temperature readings across different contexts.
Linear Equations
Linear equations are mathematical statements that describe a straight line when plotted on a graph. They typically involve variables like \( x \) or \( C \) in our case, and constants. A linear equation has the form \( ax + b = 0 \), where \( a \) and \( b \) are constants. In the context of temperature, if we equate Fahrenheit and Celsius to form a linear equation like \( F = \frac{9}{5}C + 32 \), any solution of this equation provides specific temperatures on both scales. Linear equations like these can define not just temperature relationships, but many other scientific phenomena as well. The exploration of these relationships helps us understand and predict outcomes based on mathematical reasoning.
Solving Equations with Fractions
Solving equations with fractions can initially seem complex, but it becomes straightforward with the right approach. Fractions often appear in conversions, just like in the temperature conversion formula. To solve such equations:
  • Eliminate the fractions by finding a common denominator or multiplying through by a factor that will cancel out the denominators.
  • Once the fractions are eliminated, the equation becomes simpler, and standard algebraic techniques can be applied to isolate the variable.
For example, in the equation \( \frac{10}{5}C - \frac{9}{5}C = 32 \), finding a common denominator (like 5 in this case) allows for combining terms. Simplifying the equation and solving for \( C \) helps find the temperature conversion precisely. Fraction-based equations, when solved step-by-step, become clear and depict the true arithmetic nature behind scientific relationships.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a linear \(X\) temperature scale, water freezes at \(-125.0^{\circ} \mathrm{X}\) and boils at \(375.0^{\circ} \mathrm{X}\). On a linear \(\mathrm{Y}\) temperature scale, water freezes at \(-70.00^{\circ} \mathrm{Y}\) and boils at \(-30.00^{\circ} \mathrm{Y}\). A temperature of \(50.00^{\circ} \mathrm{Y}\) corresponds to what temperature on the \(X\) scale?

The giant hornet Vespa mandarinia japonica preys on Japanese bees. However, if one of the hornets attempts to invade a beehive, several hundred of the bees quickly form a compact ball around the hornet to stop it. They don't sting, bite, crush,or suffocate it. Rather they overheat it by quickly raising their body temperatures from the normal \(35^{\circ} \mathrm{C}\) to \(47^{\circ} \mathrm{C}\) or \(48^{\circ} \mathrm{C}\), which is lethal to the hornet but not to the bees (Fig. \(18-44)\). Assume the following: 500 bees form a ball of radius \(R=2.0 \mathrm{~cm}\) for a time \(t=\) 20 min, the primary loss of energy by the ball is by thermal radiation, the ball's surface has emissivity \(\varepsilon=0.80\), and the ball has a uniform temperature. On average, how much additional energy must each bee produce during the 20 min to maintain \(47^{\circ} \mathrm{C}\) ?

An insulated Thermos contains \(130 \mathrm{~cm}^{3}\) of hot coffee at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

A room is lighted by four \(100 \mathrm{~W}\) incandescent lightbulbs. (The power of \(100 \mathrm{~W}\) is the rate at which a bulb converts electrical energy to heat and the energy of visible light.) Assuming that \(73 \%\) of the energy is converted to heat, how much heat does the room receive in \(6.9 \mathrm{~h}\) ?

A sample of gas expands from an initial pressure and volume of \(10 \mathrm{~Pa}\) and \(1.0 \mathrm{~m}^{3}\) to a final volume of \(2.0 \mathrm{~m}^{3}\). During the expansion, the pressure and volume are related by the equation \(p=a V^{2}\), where \(a=10 \mathrm{~N} / \mathrm{m}^{8}\). Determine the work done by the gas during this expansion.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.