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Chapter 12: Problem 1

Suppose you are hiking down the Grand Canyon. At the top, the temperature early in the morning is a cool \(3{ }^{\circ} \mathrm{C}\). By late afternoon, the temperature at the bottom of the canyon has warmed to a sweltering \(34^{\circ} \mathrm{C} .\) What is the difference between the higher and lower temperatures in (a) Fahrenheit degrees and (b) kelvins?

Short Answer

Expert verified
The temperature difference is 55.8掳F and 31 K.

Step by step solution

01

Temperature Conversion to Fahrenheit

First, convert both temperatures from Celsius to Fahrenheit. Use the conversion formula for Celsius to Fahrenheit: \[ F = \frac{9}{5}C + 32 \]For the top of the canyon, calculate: \[ F_{top} = \frac{9}{5} \times 3 + 32 = 37.4^{\circ} F \]For the bottom of the canyon, calculate: \[ F_{bottom} = \frac{9}{5} \times 34 + 32 = 93.2^{\circ} F \]
02

Calculate Temperature Difference in Fahrenheit

Subtract the Fahrenheit temperature at the top from the temperature at the bottom to find the difference:\[ \Delta F = F_{bottom} - F_{top} = 93.2^{\circ} F - 37.4^{\circ} F = 55.8^{\circ} F \]
03

Understanding Kelvin Conversion

To convert the temperature difference to Kelvin, we first need to know that a change of 1 degree Celsius is equivalent to a change of 1 Kelvin. This means that the difference in Celsius directly translates to the difference in Kelvin.
04

Calculate Temperature Difference in Celsius

Subtract the Celsius temperature at the top from the temperature at the bottom to find the difference:\[ \Delta C = 34^{\circ} C - 3^{\circ} C = 31^{\circ} C \]
05

Translate Celsius Difference to Kelvin Difference

Use the fact that a temperature difference of 1 degree Celsius corresponds to a 1 Kelvin change. Thus,\[ \Delta K = \Delta C = 31 \text{ K} \]

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Key Concepts

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

Celsius to Fahrenheit conversion
When converting temperatures from Celsius to Fahrenheit, there is a simple and reliable formula you can use:
  • \[ F = \frac{9}{5}C + 32 \]
The formula accounts for two main parts:
  • The factor \( \frac{9}{5} \) represents the ratio between the Celsius and Fahrenheit degrees because the Fahrenheit scale has a wider range of temperatures per degree.
  • Add 32 to adjust the starting point of the two scales, as 0掳C corresponds to 32掳F.
Let's take a closer look at our example:
For a temperature at the top of the Grand Canyon, which is \( 3^{\circ} \mathrm{C} \):
  • \[ F_{\text{top}} = \frac{9}{5} \times 3 + 32 = 37.4^{\circ} F \]
At the bottom with \( 34^{\circ} \mathrm{C} \), we calculate:
  • \[ F_{\text{bottom}} = \frac{9}{5} \times 34 + 32 = 93.2^{\circ} F \]
This conversion is important when you want to express temperatures in a different scale, such as when travelling to countries using the opposite system.
temperature difference
Understanding temperature differences involves subtracting one temperature from another, which helps describe the change or range.
Here鈥檚 how it works in both Celsius and Fahrenheit:
  • First, find the difference in Celsius: \[ \Delta C = 34^{\circ} C - 3^{\circ} C = 31^{\circ} C \]
  • Then convert these temperatures to Fahrenheit as calculated before: find the difference in Fahrenheit:\[ \Delta F = 93.2^{\circ} F - 37.4^{\circ} F = 55.8^{\circ} F \]
Calculating temperature differences is quite useful:
  • It helps understand how much the temperature varied over the day or between locations.
  • Important for planning outdoor activities, as seen in hiking, where temperature changes can significantly impact comfort and safety.
Kelvin scale
The Kelvin scale is a thermodynamic (absolute) temperature scale often used in scientific contexts.
Understanding the Kelvin scale:
  • Zero Kelvin (0 K) is absolute zero, the point where molecular motion stops.
  • Unlike Celsius and Fahrenheit, the Kelvin scale does not use degrees and starts at absolute zero.
  • One Kelvin is exactly the same size as one degree Celsius, meaning they change in parallel.
When calculating temperature differences with Kelvin, the method is straightforward:
  • Directly use the Celsius difference: \[ \Delta K = \Delta C = 31 \text{ K} \]
The Kelvin scale is crucial for scientists because it provides a comprehensive and universally consistent scale for measuring temperature differences, not subject to the subjective starting points like Celsius and Fahrenheit.

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Most popular questions from this chapter

A copper-constantan thermocouple generates a voltage of \(4.75 \times\) \(10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of \(0.0^{\circ} \mathrm{C} .\) If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

Water at \(23.0^{\circ} \mathrm{C}\) is sprayed onto \(0.180 \mathrm{kg}\) of molten gold at \(1063^{\circ} \mathrm{C}\) (its melting point). The water boils away, forming steam at \(100.0^{\circ} \mathrm{C}\) and leaving solid gold at \(1063^{\circ} \mathrm{C} .\) What is the minimum mass of water that must be used?

A 42 -kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{m} / \mathrm{s} .\) Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

At the bottom of an old mercury-in-glass thermometer is a \(45-\mathrm{mm}^{3}\) reservoir filled with mercury. When the thermometer was placed under your tongue, the warmed mercury would expand into a very narrow cylindrical channel, called a capillary, whose radius was \(1.7 \times 10^{-2} \mathrm{mm}\). Marks were placed along the capillary that indicated the temperature. Ignore the thermal expansion of the glass and determine how far (in \(\mathrm{mm}\) ) the mercury would expand into the capillary when the temperature changed by \(1.0 \mathrm{C}^{\circ}\)

The vapor pressure of water at \(10^{\circ} \mathrm{C}\) is \(1300 \mathrm{Pa}\). (a) What percentage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5}\) Pa. (b) What percentage of the total air pressure at \(10^{\circ} \mathrm{C}\) is due to water vapor when the relative humidity is \(100 \% ?\) (c) The vapor pressure of water at \(35^{\circ} \mathrm{C}\) is 5500 Pa. What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} \mathrm{C}\) when the relative humidity was \(100 \% ?\)
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