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\(\bullet\) (a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees and in kelvins? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{Fon}\) January \(23,1916,\) and the next day it plummeted to \(-56.0^{\circ} \mathrm{F} .\) What was the temperature change in Celsius degrees and in kelvins?

Short Answer

Expert verified
(a) 27.2°C, 27.2 K; (b) -55.6°C, 55.6 K.

Step by step solution

01

Calculate Temperature Change in Fahrenheit for Part (a)

To find the change in temperature in Fahrenheit for the first situation, subtract the initial temperature from the final temperature: \[ 45.0^{\circ} \text{F} - (-4.0^{\circ} \text{F}) = 45.0^{\circ} \text{F} + 4.0^{\circ} \text{F} = 49.0^{\circ} \text{F} \]
02

Convert Fahrenheit Change to Celsius for Part (a)

Use the conversion factor between Fahrenheit and Celsius: \[ \Delta T(^{\circ}C) = \frac{\Delta T(^{\circ}F)}{1.8} \]Substitute the Fahrenheit change obtained in Step 1: \[ \Delta T(^{\circ}C) = \frac{49.0}{1.8} = 27.2^{\circ}C \]
03

Convert Celsius Change to Kelvins for Part (a)

Since changes in temperature are the same in both Celsius and Kelvin (1 degree Celsius change is equivalent to 1 Kelvin change), the change in Kelvin is:\[ \Delta T(K) = 27.2 K \]
04

Calculate Temperature Change in Fahrenheit for Part (b)

For the second situation, calculate the change in Fahrenheit: \[ -56.0^{\circ} \text{F} - 44.0^{\circ} \text{F} = -56.0^{\circ} \text{F} + (-44.0^{\circ} \text{F}) = -100.0^{\circ} \text{F} \]
05

Convert Fahrenheit Change to Celsius for Part (b)

Use the same conversion as in Step 2: \[ \Delta T(^{\circ}C) = \frac{-100.0}{1.8} = -55.6^{\circ}C \]
06

Convert Celsius Change to Kelvins for Part (b)

Like in Step 3, the change in Kelvin is the same as the change in Celsius:\[ \Delta T(K) = 55.6 K \]

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Key Concepts

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

Temperature Change
Temperature change refers to the difference in temperature between two points in time. In the case of the exercise, we are asked to calculate this change when temperatures fluctuate in both Fahrenheit and then convert this change into Celsius and Kelvin.
  • Spearfish, South Dakota: The temperature increase was from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\).
  • Browning, Montana: The temperature decrease was from \(44.0^{\circ} \mathrm{F}\) to \(-56.0^{\circ} \mathrm{F}.\)
To compute the change, subtract the initial temperature from the final temperature in Fahrenheit. It's important to consider the signs since they show whether the change is an increase or a decrease. Calculating in Fahrenheit first helps because it is often the unit used in the problem. A positive result means an increase in temperature, while a negative result indicates a decrease.
Understanding temperature change is crucial, as it helps to interpret weather events, like rapid temperature changes, which may occur due to climatic conditions.
Fahrenheit to Celsius Conversion
Converting from Fahrenheit to Celsius requires using the formula that accounts for the different starting points of the two scales. The formula for converting a Fahrenheit temperature change to a Celsius temperature change is given by:\[\Delta T(^{\circ}C) = \frac{\Delta T(^{\circ}F)}{1.8}\]To apply the formula:
  • Divide the change in Fahrenheit by \(1.8\).
  • This conversion factor comes from the relationship that \(1^{\circ}\mathrm{C} = 1.8^{\circ}\mathrm{F}.\)
For instance, in part (a) of the exercise, a \(49.0^{\circ}\mathrm{F}\) change is converted to \(27.2^{\circ}\mathrm{C}\), while in part (b), a \(-100.0^{\circ}\mathrm{F}\) change becomes \(-55.6^{\circ}\mathrm{C}.\)
Understanding this conversion is vital in science, especially in contexts where Celsius is often more commonly used, such as chemistry and physics.
Celsius to Kelvin Conversion
While converting a change in temperature from Celsius to Kelvin, students often find that it's actually quite direct and simple. The Kelvin scale is essential in scientific calculations because it starts at absolute zero, the lowest possible temperature where particles have minimum thermal energy.
  • **Add 273.15**: While this is necessary for converting a specific temperature to Kelvin (e.g., 0°C is 273.15 K), **no addition** is necessary for a temperature change.
  • **1°C change = 1 K change**: This is because both Celsius and Kelvin have the same increment size.
In the provided exercise, we calculated the change in Celsius and directly translated that to Kelvin.
So, a \(27.2^{\circ}C\) change is equivalent to a \(27.2\mathrm{K}\) change, and similarly, \(-55.6^{\circ}C\) is simply \(55.6\mathrm{K}\).
This simplicity is particularly useful in thermal physics, where Kelvin is prevalent due to its direct link with the fundamental properties of particles.

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Most popular questions from this chapter

\(\cdot\) The Eiffel Tower in Paris is 984 ft tall and is made mostly of steel. If this is its height in winter when its temperature is \(-8.00^{\circ} \mathrm{C},\) how much additional vertical distance must you cover if you decide to climb it during a summer heat wave when its temperature is \(40.0^{\circ} \mathrm{C}\) ? (b) Express the coefficient of linear expansion of steel in terms of Fahrenheit degrees.

. A thermos for liquid helium. A physicist uses a cylindrical metal can 0.250 \(\mathrm{m}\) high and 0.090 \(\mathrm{m}\) in diameter to store liquid helium at \(4.22 \mathrm{K} ;\) at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 \(\mathrm{K}\) , with vacuum between the can and the surrounding walls. How much helium is lost per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.

\(\bullet\) The effect of urbanization on plant growth. A study published in July 2004 indicated that temperature increases in urban areas in the eastern United States are causing plants to bud up to 7 days early compared with plants in rural areas just a few miles away, thereby disrupting biological cycles. Average temperatures in the urban areas were up to 3.5 \(\mathrm{C}^{\circ}\) higher than in the rural areas. By what percent will the radiated heat per square meter increase due to such a temperature difference if the rural temperature was \(0^{\circ} \mathrm{C}\) the average?

. An ice-cube tray contains 0.350 \(\mathrm{kg}\) of water at \(18.0^{\circ} \mathrm{C}\) . How much heat must be removed from the water to cool it to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules and in calories.

Maintaining body temperature. While running, a 70 \(\mathrm{kg}\) student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are damaged irreversibly if the body temperature rises to \(44^{\circ} \mathrm{C}\) or above. The specific heat capacity of a typical human body is \(3480 \mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}),\) slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heat capacities.)

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