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Starting from \(T_{\mathrm{C}}=\frac{5}{9}\left(T_{\mathrm{F}}-32\right)\) (Equation 14-2), derive a formula for converting from \({ }^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\). Why is it more common for the multiplicative factor in Equation 14-2 to be written as a fraction rather than a decimal?

Short Answer

Expert verified
The formula to convert temperature from Celsius to Fahrenheit is \(T_{\mathrm{F}} = \frac{9}{5} \cdot T_{\mathrm{C}} + 32\). The fraction form is more common because it represents a more exact value and avoids potential rounding errors that can occur with decimal representation.

Step by step solution

01

Write Down the Given Equation and Understand its Components

The current equation represents Fahrenheit in terms of Celsius: \(T_{\mathrm{C}}=\frac{5}{9}\left(T_{\mathrm{F}}-32\right)\). In this equation, \(T_C\) is the temperature in Celsius and \(T_F\) is the temperature in Fahrenheit. Ultimately, the goal is to express \(T_F\) in terms of \(T_C\).
02

Isolate \(T_{\mathrm{F}}\)

To isolate \(T_{\mathrm{F}}\), the first step is to multiply both sides by \(\frac{9}{5}\). By doing this, we obtain \(\frac{9}{5} \cdot T_{\mathrm{C}} = T_{\mathrm{F}} - 32\). Then, to completely isolate \(T_{\mathrm{F}}\), add 32 to both sides of the equation. This yields the equation \(T_{\mathrm{F}} = \frac{9}{5} \cdot T_{\mathrm{C}} + 32\). This equation gives the temperature in Fahrenheit as a function of the temperature in Celsius.
03

Reason for Fractional Format

The second part of the problem asks why we use a fraction, \(\frac{5}{9}\), in the conversion instead of a decimal. This is usually because a fractional form provides a more exact representation in mathematical equations. For instance, the exact value of \(\frac{5}{9}\) is about 0.55556 to five decimal places. Writing this as a decimal would potentially introduce rounding errors, particularly in precise scientific calculations. By keeping the coefficient in fractional form, this issue is avoided.

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Key Concepts

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

Celsius to Fahrenheit
Converting from Celsius to Fahrenheit is a common task in science and daily life. We start with an equation that connects these two temperature scales: \(T_{\mathrm{C}} = \frac{5}{9}\left(T_{\mathrm{F}}-32\right)\). The goal is to express the Fahrenheit temperature \(T_F\) in terms of Celsius \(T_C\).

By manipulating the equation, we can find how to convert Celsius to Fahrenheit. First, multiply both sides by \(\frac{9}{5}\) to get rid of the fraction on the left-hand side. This results in \(\frac{9}{5} \cdot T_{\mathrm{C}} = T_{\mathrm{F}} - 32\). Adding 32 to both sides isolates \(T_{\mathrm{F}}\):
- \(T_{\mathrm{F}} = \frac{9}{5} \cdot T_{\mathrm{C}} + 32)\)

This equation shows clearly how every degree Celsius corresponds to an increase of \(1.8\) degrees Fahrenheit after considering the 32-degree offset. This conversion is handy when you need to comprehend everyday weather conditions or scientific data, as many countries use Celsius while others use Fahrenheit.
Fahrenheit to Celsius
Converting from Fahrenheit to Celsius utilizes the reverse of the logic we used for Celsius to Fahrenheit. Given the equation \(T_{\mathrm{C}} = \frac{5}{9}\left(T_{\mathrm{F}}-32\right)\), we see how Fahrenheit temperatures convert to Celsius ones.

The first step is to subtract 32 from the Fahrenheit temperature to account for the difference in the zero points of the two scales. For example, water freezes at 32°F which is equivalent to 0°C. Then, multiply this difference by \(\frac{5}{9}\).

This equation, \(T_{\mathrm{C}} = \frac{5}{9}(T_{\mathrm{F}}-32)\), helps many in understanding global weather reports or preparing food according to international recipes. It emphasizes how a Fahrenheit temperature needs to be scaled down, considering that a Fahrenheit degree is smaller than a Celsius degree.
Fractional Precision
Fractions often offer more precision than their decimal counterparts, making them preferable in scientific equations. Consider \(\frac{5}{9}\), which is used in converting Fahrenheit to Celsius. Its decimal approximation is \(0.55556\), accurate to five decimal places.

Using fractions eliminates the rounding errors that decimals might introduce. When calculations require high precision, even minute errors from decimals can significantly affect the outcome. For instance, this precision is crucial in scientific experiments or engineering feats where every digit contributes to the reliability of results.

Fractions also communicate exact values to those who understand them. While less intuitive at first glance than a decimal, \(\frac{5}{9}\) straightforwardly reveals that each part of the fraction - 5 and 9 - is critical to maintaining the integrity of the conversion. This meticulous attention to detail exemplifies why fractional precision remains vital in many mathematical and scientific contexts.

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Most popular questions from this chapter

Biology A \(1.88 \mathrm{~m}\) (6 ft \(2 \mathrm{in}\).) man has a mass of \(80 \mathrm{~kg}\), a body surface area of \(2.1 \mathrm{~m}^{2}\) and a skin temperature of \(30{ }^{\circ} \mathrm{C}\). Normally \(80 \%\) of the food calories he consumes go to heat, the rest going to mechanical energy. To keep his body's temperature constant, how many food calories should he eat per day if he is in a room at \(20^{\circ} \mathrm{C}\) and he loses heat only through radiation? Does the answer seem reasonable? His emissivity \(\varepsilon\) is 1 because his body radiates almost entirely nonvisible infrared energy, which is not affected by skin pigment. (Careful! His body at \(30^{\circ} \mathrm{C}\) radiates into the air at \(20^{\circ} \mathrm{C}\), but the air also radiates back into his body. The net rate of radiation is \(P_{\text {nat }}=P_{\text {bodr }}-P_{\text {air }-\text { ) SSM }}\)

Explain the term latent as it applies to phase changes.

Convert the following remperatures and comment on any physical significance of each: A. \(0^{\circ} \mathrm{C}=\) B. \(212{ }^{\circ} \mathrm{F}=\) C. \(273 \mathrm{~K}=\) D. \(68^{\circ} \mathrm{F}=\)

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Biology You may have noticed that small mammals (such as mice) seem to be constantly eating, whereas some large mammals (such as lions) eat much less frequently. Let us investigate the phenomenon. For simplicity we can model an animal as a sphere. (a) Show that the heat energy stored by an animal is proportional to the cube of its radius, but the rate at which the animal radiates energy away is proportional to the square of its radius. (b) Show that the fraction of the animal's stored energy that it radiates away per second is inversely proportional to the animal's radius. (c) Use the result in part (b) to explain why small animals must eat much more per gram of body weight than very large animals.

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