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Chapter 10: Problem 25

At what temperature, the Fahrenheit and the Celsius scales will give numerically equal (but opposite in sign) values? (A) \(-40^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{C}\) (B) \(11.43^{\circ} \mathrm{F}\) and \(-11.43{ }^{\circ} \mathrm{C}\) (C) \(-11.43^{\circ} \mathrm{F}\) and \(+11.43{ }^{\circ} \mathrm{C}\) (D) \(+40^{\circ} \mathrm{F}\) and \(-40{ }^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature at which the Fahrenheit and Celsius scales give numerically equal but opposite in sign values is \(+40^{\circ} \mathrm{F}\) and \(-40{ }^{\circ} \mathrm{C}\).

Step by step solution

01

Write the formula to convert Celsius to Fahrenheit

To convert from Celsius to Fahrenheit, we use the following formula: \[ F = \frac{9}{5}C + 32 \]
02

Set up the equation to find the equal but opposite values

Since we are looking for the temperature where the Fahrenheit (F) and Celsius (C) values are numerically equal but opposite in sign, we can set up the following equation: \[ F = -C \]
03

Solve for the temperature in terms of Celsius

Plug in the value of F from the equation (F = -C) into the Fahrenheit to Celsius conversion formula (F = 9/5C + 32) and solve for C: \[ -C = \frac{9}{5}C + 32 \] Add C to both sides: \[ 0 = \left(\frac{9}{5}+1\right)C + 32 \] With \( \frac{9}{5}+1=\frac{14}{5}\), we get: \[ 0 = \frac{14}{5}C + 32 \] Now, subtract 32 from both sides: \[ -32 = \frac{14}{5}C \]
04

Solve for the temperature in Celsius

To find the value of C, we will divide both sides by (14/5): \[ C = \frac{-32}{\frac{14}{5}} \] To divide by a fraction, we multiply by its reciprocal, i.e., \( \frac{5}{14}\): \[ C = -32 \times \frac{5}{14} \] Simplifying, we get: \[ C = -40 \]
05

Find the temperature in Fahrenheit

Now that we have the temperature in Celsius, we can find the temperature in Fahrenheit using the Fahrenheit to Celsius conversion formula (F = 9/5C + 32): \[ F = \frac{9}{5}(-40) + 32 \] Simplifying, we get: \[ F = -72 + 32 = -40 \] So, the answer is (D) \(+40^{\circ} \mathrm{F}\) and \(-40{ }^{\circ} \mathrm{C}\).

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Key Concepts

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

Fahrenheit to Celsius conversion
Understanding how to convert from Fahrenheit to Celsius is a foundational skill in physics, especially when studying temperature-related phenomena. The formula to convert Celsius to Fahrenheit is represented as:
\[ F = \frac{9}{5}C + 32 \]
This equation implies that for every degree Celsius, it translates to 1.8 degrees Fahrenheit, and then 32 degrees are added to adjust for the offset between the two scales at the point where they both read zero.For instance, the freezing point of water in Celsius is 0 degrees, but in Fahrenheit, it is 32 degrees because when you plug 0 into the formula, it translates to 32 degrees Fahrenheit. Conversely, to convert from Fahrenheit to Celsius, the equation would be rearranged to solve for C:
\[ C = \frac{5}{9}(F - 32) \]
These conversions are crucial in several real-world applications, from cooking recipes to climate data analyses and scientific experiments.
Thermodynamics problem solving
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. Problem-solving in thermodynamics often involves understanding and manipulating various equations and concepts to find unknown variables.When tackling a thermodynamics problem, such as finding a specific temperature when scales converge, a systematic approach must be taken. This includes clearly defining all knowns and unknowns, choosing the relevant formulas, substituting known values into these equations, and doing the arithmetic carefully. Such a structured method not only helps in arriving at the correct answer but also ingrains the principle behind the process.In the case of the exercise provided, by merging the concept of temperature scale conversion with an understanding of algebraic manipulation, the solution becomes apparent. It's a good reminder that even within the complex field of thermodynamics, sometimes the problems we face can be simplified and solved with basic algebraic techniques.
JEE Main Physics
Physics is a pivotal subject in the Joint Entrance Examination (JEE) Main, an entrance test for various engineering colleges in India. The JEE Main Physics section tests students on their conceptual understanding and problem-solving abilities across a range of topics, including mechanics, electromagnetism, and thermodynamics.Success in JEE Main Physics requires a deep understanding of concepts and proficiency in their application through various types of problems. Conversion of temperature scales, while seeming like a simple topic, is a fundamental concept that can be applied to more complex thermodynamic problems. It's important for students to practice these types of exercises regularly as it helps solidify their grasp of the concepts, develops their problem-solving skills, and increases their speed and accuracy - all of which are crucial for performing well in competitive exams like the JEE Main. 91影视 and study materials that break down difficult ideas into easy-to-understand content, such as clear, step-by-step calculations, can significantly benefit JEE aspirants.

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Most popular questions from this chapter

An aluminium sphere is dipped into water. Which of the following is true? (A) Buoyancy will be less in water at \(0^{\circ} \mathrm{C}\) than that in water at \(4^{\circ} \mathrm{C}\) (B) Buoyancy will be more in water at \(0^{\circ} \mathrm{C}\) than that in water at \(4^{\circ} \mathrm{C}\) (C) Buoyancy in water at \(0^{\circ} \mathrm{C}\) will be same as that in water at \(4^{\circ} \mathrm{C}\) (D) Buoyancy may be more or less in water at \(4^{\circ} \mathrm{C}\) depending on the radius of the sphere

The ratio of coefficients of cubical expansion and linear expansion is (A) \(1: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) None of these

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\) If the difference between the two lengths is independent of temperature (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)

A solid body of constant heat capacity \(1 \mathrm{~J} /{ }^{\circ} \mathrm{C}\) is being heated by keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the case body is brought from initial temperature \(100^{\circ} \mathrm{C}\) to final temperature \(200^{\circ} \mathrm{C}\). Entropy change of the body in the two cases respectively is \([2015]\) (A) \(\ln 2, \ln 2\) (B) \(\ln 2,2 \ln 2\) (C) \(2 \ln 2,8 \ln 2\) (D) \(\ln 2,4 \ln 2\)

Gulab jamuns (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice bigger (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following. (A) Both size gulab jamuns will get heated in the same time. (B) Smaller gulab jamuns are heated before bigger ones. (C) Smaller pizzas are heated before bigger ones. (D) Bigger pizzas are heated before smaller.
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