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(I) (a) "Room temperature" is often taken to be \(68^{\circ} \mathrm{F}\). What is this on the Celsius scale? (b) The temperature of the filament in a lightbulb is about \(1900^{\circ} \mathrm{C}\). What is this on the Fahrenheit scale?

Short Answer

Expert verified
(a) 68°F is 20°C. (b) 1900°C is 3452°F.

Step by step solution

01

Understand the Conversion Formula

To convert temperatures between Celsius and Fahrenheit, use the formula:\[C = \frac{5}{9}(F - 32)\]where \(C\) is the temperature in degrees Celsius and \(F\) is the temperature in degrees Fahrenheit.
02

Convert Fahrenheit to Celsius

We are given that room temperature is \(68^{\circ} \mathrm{F}\). To find this in Celsius, we use the conversion formula:\[C = \frac{5}{9}(68 - 32)\]Simplifying this:\[C = \frac{5}{9} \times 36 = 20^{\circ} \mathrm{C}\]
03

Understand the Conversion Formula for Celsius to Fahrenheit

To convert temperatures between Celsius and Fahrenheit, the formula is:\[F = \frac{9}{5}C + 32\]where \(C\) is the temperature in degrees Celsius and \(F\) is the temperature in degrees Fahrenheit.
04

Convert Celsius to Fahrenheit

The temperature of the filament is \(1900^{\circ} \mathrm{C}\). To convert this to Fahrenheit, use the formula:\[F = \frac{9}{5} \times 1900 + 32\]Simplifying this:\[F = 3420 + 32 = 3452^{\circ} \mathrm{F}\]

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Key Concepts

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

Celsius to Fahrenheit
When converting temperatures from Celsius to Fahrenheit, it's important to use the conversion formula: \[ F = \frac{9}{5}C + 32 \]Here, \(C\) represents the temperature in degrees Celsius, which you want to convert into the Fahrenheit scale, represented by \(F\). This formula essentially scales the Celsius temperature by \(\frac{9}{5}\), representing the larger size of each Fahrenheit degree, and then adds 32. The added 32 is crucial as it adjusts the zero point from Celsius to Fahrenheit. For example, if you have a temperature of \(1900^{\circ} C\) you can convert it to Fahrenheit by substituting into the formula:\[ F = \frac{9}{5} \times 1900 + 32 \]When you carry out the multiplication and addition, you will find that 1900 degrees Celsius equals \(3452^{\circ} F\). This conversion is particularly useful in scientific settings or when dealing with high temperatures, as Fahrenheit might be used in different contexts, especially in the U.S.
Fahrenheit to Celsius
To convert temperatures from Fahrenheit to Celsius, you can use the formula:\[ C = \frac{5}{9}(F - 32) \]Here, \(F\) is the temperature in Fahrenheit that needs to be converted to Celsius, represented by \(C\). The formula first requires you to subtract 32 from the Fahrenheit temperature, because the freezing point of water is at 32°F not 0°C. Then, the value is scaled by \(\frac{5}{9}\) since each Celsius degree is larger.Suppose you have a Fahrenheit temperature of \(68^{\circ}F\). Plugging this into the formula gives:\[ C = \frac{5}{9} (68 - 32) \]Simplifying this results in \(C = 20^{\circ}C\). This conversion step is most common in everyday settings where Fahrenheit is used, but the Celsius scale is required for scientific or international contexts.
Temperature Scales
Temperature scales are frameworks for measuring temperature, with the most common ones being Celsius and Fahrenheit.
  • Celsius (\(^{\circ}C\): The Celsius scale is used worldwide and sets the freezing point of water at 0 degrees and the boiling point at 100 degrees under normal atmospheric pressure. It's often used in scientific measurements due to its decimal-based nature.
  • Fahrenheit (\(^{\circ}F\): Predominantly used in the United States, Fahrenheit sets the freezing point of water at 32 degrees and the boiling point at 212 degrees. It results in temperatures that are typically higher numbers than Celsius for the same physical situational measurement.
Understanding the differences in these scales is important for making accurate temperature conversions. Each scale provides useful information depending on the context—be it scientific, geographic, or cultural.

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Most popular questions from this chapter

A house has a volume of 870 \(\mathrm{m}^{3} .\) (a) What is the total mass of air inside the house at \(15^{\circ} \mathrm{C} ?\) (b) If the temperature drops to \(-15^{\circ} \mathrm{C}\) , what mass of air enters or leaves the house?

(II) A thermocouple consists of a junction of two different types of materials that produces a voltage depending on its temperature. A thermocouple's voltages were recorded when at different temperatures as follows: \begin{tabular}{lrrrr} \hline Temperature \(\left({ }^{\circ} \mathrm{C}\right)\) & 50 & 100 & 200 & 300 \\ Voltage \((\mathrm{mV})\) & 1.41 & 2.96 & 5.90 & 8.92 \\ \hline \end{tabular} Use a spreadsheet to fit these data to a cubic equation and determine the temperature when the thermocouple produces \(3.21 \mathrm{mV}\). Get a second value of the temperature by fitting the data to a quadratic equation.

(II) A stoppered test tube traps \(25.0 \mathrm{~cm}^{3}\) of air at a pressure of 1.00 atm and temperature of \(18^{\circ} \mathrm{C}\). The cylindrically shaped stopper at the test tube's mouth has a diameter of \(1.50 \mathrm{~cm}\) and will "pop off" the test tube if a net upward force of \(10.0 \mathrm{~N}\) is applied to it. To what temperature would one have to heat the trapped air in order to "pop off' the stopper? Assume the air surrounding the test tube is always at a pressure of 1.00 atm.

(II) In an alcohol-in-glass thermometer, the alcohol column has length \(11.82 \mathrm{~cm}\) at \(0.0^{\circ} \mathrm{C}\) and length \(21.85 \mathrm{~cm}\) at \(100.0^{\circ} \mathrm{C}\). What is the temnerature if the column has leneth \((a) 18.70 \mathrm{~cm}\). and

Assume that in an alternate universe, the laws of physics are very different from ours and that "ideal" gases behave as follows: (i) At constant temperature, pressure is inversely proportional to the square of the volume. (ii) At constant pressure, the volume varies directly with the \(\frac{2}{3}\) power of the temperature. (iii) At \(273.15 \mathrm{~K}\) and 1.00 atm pressure, 1.00 mole of an ideal gas is found to occupy \(22.4 \mathrm{~L}\).

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