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(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? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{F}\). What was the temperature change in Celsius degrees?

Short Answer

Expert verified
The temperature change in Celsius degrees was \(9.444^{\circ} \mathrm{C}\) for the event in Spearfish, South Dakota, and \(37.778^{\circ} \mathrm{C}\) for the event in Browning, Montana.

Step by step solution

01

Calculate the temperature difference in Fahrenheit for the first event

In Spearfish, South Dakota, the temperature rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\), so the temperature difference in Fahrenheit is \(45.0^{\circ} \mathrm{F} - (-4.0^{\circ} \mathrm{F}) = 49.0^{\circ} \mathrm{F}\).
02

Convert the temperature difference to Celsius for the first event

Having calculated the temperature difference in Fahrenheit, it can now be converted to Celsius using the formula \(C = \frac{5}{9}(F - 32)\). Substituting the given value for \(F\), we obtain \( C = \frac{5}{9}(49.0^{\circ} \mathrm{F} - 32) = 9.444^{\circ} \mathrm{C}\).
03

Calculate the temperature difference in Fahrenheit for the second event

In Browning, Montana, the temperature dropped from \(44.0^{\circ} \mathrm{F}\) to \(-56^{\circ} \mathrm{F}\), so the temperature difference in Fahrenheit is \(44.0^{\circ} \mathrm{F} - (-56^{\circ} \mathrm{F}) = 100.0^{\circ} \mathrm{F}\).
04

Convert the temperature difference to Celsius for the second event

Using the same formula to convert temperature from Fahrenheit to Celsius, we obtain \( C = \frac{5}{9}(100.0^{\circ} \mathrm{F} - 32) = 37.778^{\circ} \mathrm{C}\).

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Key Concepts

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

Fahrenheit to Celsius Conversion
The temperature scale can sometimes puzzle students, as different countries use different systems. In the United States, the Fahrenheit scale is the standard, while most of the world uses Celsius. Understanding how to convert from Fahrenheit to Celsius is vital, especially in science classes and for international communication.

The formula to convert Fahrenheit to Celsius is surprisingly simple: \( C = \frac{5}{9}(F - 32) \), where \(C\) represents degrees Celsius and \(F\) represents degrees Fahrenheit. This formula is derived from the different boiling and freezing points of water on the two scales: 32°F is the freezing point of water, which corresponds to 0°C, and 212°F is the boiling point of water, which corresponds to 100°C. The subtraction of 32 adjusts for the difference in freezing points, while the multiplication by \(\frac{5}{9}\) adjusts for the different scales of degrees between the systems.

To envision this practically: if you have a temperature of 50°F, applying the conversion formula would give you \( C = \frac{5}{9}(50 - 32) = 10\degree C \). This process of conversion is fundamental in understanding temperature differences and conducting accurate science experiments.
Temperature Change Calculation
When we talk about temperature change, we're referring to the difference between two temperature measurements, taken at different times or under different conditions. In mathematics and physics, this is often represented as a simple subtraction problem. However, it's crucial to remember that this calculation depends on the scale being used.

For example, the rapid temperature change in Spearfish, South Dakota, can be mathematically represented by subtracting the initial temperature from the final temperature: \( 45.0^\circ F - (-4.0^\circ F) = 49.0^\circ F \). Such a calculation shows us the literal change in temperature, but to fully understand the scale of this change, converting it into Celsius provides a more uniform understanding globally.

The conversion to Celsius gives us the temperature change as \( 9.444^\circ C \), which may be more relatable if you're accustomed to the metric system. Understanding how to perform this calculation is important for scientists, meteorologists, and anyone interested in the quantitative aspects of temperature changes.
Temperature Difference in Physics
The temperature difference isn't just a number; it has significant implications in the field of physics, particularly in the study of thermodynamics. The temperature difference can indicate the flow of heat from one substance to another and is critical in understanding physical processes like the efficiency of engines and refrigerators, or the weather phenomena that occur due to sudden temperature changes like the one in Spearfish, South Dakota.

In physics, temperature difference can impact the behavior of materials, such as the expansion and contraction of metals or the pressure of gases. Seeing the temperature plunge from \(44.0^\circ F\) to a frigid \(-56^\circ F\), as in Browning, Montana, represents a dramatic shift in environmental conditions. Such shifts can cause physical stress to materials and living organisms, leading to interesting study cases in physics. The temperature difference of \(37.778^\circ C\) calculated for this event provides a tangible figure for such a study.

Understanding temperature difference in this scientific context allows students to grasp the real-world implications of temperature change, beyond mere numbers on a thermometer.

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Most popular questions from this chapter

On-Demand Water Heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is \(2.5 \mathrm{gal} / \mathrm{min}(9.46 \mathrm{~L} / \mathrm{min})\) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

An insulated beaker with negligible mass contains \(0.250 \mathrm{~kg}\) of water at \(75.0^{\circ} \mathrm{C}\). How many kilograms of ice at \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

A U.S. penny has a diameter of \(1.9000 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\). The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{~K}^{-1}\). What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a \(70.0 \mathrm{~kg}\) man to cool his body \(1.00 \mathrm{C}^{\circ}\) ? The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) The specific heat of a typical human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.27 ) . (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{~cm}^{3}\right)\).

The molar heat capacity of a certain substance varies with temperature according to the empirical equation $$ C=29.5 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+\left(8.20 \times 10^{-3} \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}^{2}\right) T $$ How much heat is necessary to change the temperature of \(3.00 \mathrm{~mol}\) of this substance from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate. \()\)

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