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Chapter 14: Problem 1

\(\bullet\) (a) While vacationing in Europe, you feel sick and are told that you have a temperature of \(40.2^{\circ} \mathrm{C}\) . Should you be concerned? What is your temperature in \(^{\circ} \mathrm{F} ?\) (b) The morning weather report in Sydney predicts a high temperature of \(12^{\circ} \mathrm{C}\) . Will you need to bring a jacket? What is this temperature in \(^{\circ} \mathrm{F} ?(\mathrm{c})\) A friend has suggested that you go swimming in a pool having water of temperature 350 \(\mathrm{K}\) . Is this safe to do? What would this temperature be on the Fahrenheit and Celsius scales?

Short Answer

Expert verified
(a) 104.36掳F indicates a fever. (b) 53.6掳F requires a jacket. (c) 350 K is dangerously hot at 170.33掳F.

Step by step solution

01

Convert Celsius to Fahrenheit for part (a)

To convert Celsius to Fahrenheit, use the formula: \[ F = \left( \frac{9}{5} \times C \right) + 32 \]Given that your temperature is 40.2掳C:\[ F = \left( \frac{9}{5} \times 40.2 \right) + 32 \]\[ F = 72.36 + 32 \]\[ F = 104.36^{\circ} \mathrm{F} \]Since a normal body temperature is around 98.6掳F, a temperature of 104.36掳F indicates a fever, and you should be concerned.
02

Convert Celsius to Fahrenheit for part (b)

Use the same conversion formula for the morning temperature in Sydney:\[ F = \left( \frac{9}{5} \times 12 \right) + 32 \]\[ F = 21.6 + 32 \]\[ F = 53.6^{\circ} \mathrm{F} \]Since 53.6掳F is quite cool, you will likely need a jacket.
03

Convert Kelvin to Celsius for part (c)

First, convert the temperature from Kelvin to Celsius using the formula:\[ C = K - 273.15 \]Given the water temperature is 350 K:\[ C = 350 - 273.15 \]\[ C = 76.85^{\circ} \mathrm{C} \]Since typical swimming pool temperatures are around 25掳C to 28掳C, 76.85掳C is dangerously hot.
04

Convert Celsius to Fahrenheit for the pool temperature

Now convert 76.85掳C (from Kelvin conversion) to Fahrenheit:\[ F = \left( \frac{9}{5} \times 76.85 \right) + 32 \]\[ F = 138.33 + 32 \]\[ F = 170.33^{\circ} \mathrm{F} \]This temperature is extremely hot for swimming, and it is not safe.

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Key Concepts

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

Celsius to Fahrenheit
Converting temperatures from Celsius to Fahrenheit is often necessary when traveling or dealing with information shared in different regions of the world. The formula to convert Celsius (C) to Fahrenheit (F) is: \[ F = \left( \frac{9}{5} \times C \right) + 32 \].
  • To use this formula, first multiply the Celsius temperature by \(\frac{9}{5}\).
  • Next, add 32 to the result to obtain the Fahrenheit temperature.
For example, a temperature recorded as 12掳C in Sydney translates to 53.6掳F based on this calculation.

Knowing this conversion is crucial, especially when assessing temperatures like 40.2掳C, which equals 104.36掳F and signifies a fever since normal body temperature is about 98.6掳F.
Kelvin to Celsius
The Kelvin to Celsius conversion is straightforward, as both scales increment at the same rate. To make the conversion, you simply need to subtract273.15from the Kelvin temperature. The formula is: \[ C = K - 273.15 \].
  • This conversion is essential in scientific settings where Kelvin is commonly used.
  • By subtracting 273.15, you translate temperatures into the more familiar Celsius scale.
For instance, a water temperature of 350 Kelvin converts to 76.85掳C.

Such a high temperature indicates it is much too hot for safe swimming. Generally, swimming pools maintain a temperature between 25掳C and 28掳C, making this conversion especially pertinent for safety assessments.
Temperature Safety
Understanding temperature safety is important for assessing the risk of certain weather conditions or environments. When temperatures fall outside safe ranges, they can pose a threat to health and safety.
  • Body temperatures above 98.6掳F (37掳C) may indicate fever or illness.
  • Environmental temperatures like a pool at 76.85掳C are extremely dangerous and unsuitable for activities.
Awareness of appropriate and safe temperature ranges is critical in both daily activities and for health monitoring.

When the temperature is predicted to be 12掳C (53.6掳F), it's advisable to prepare for cooler weather, such as bringing a jacket if going out.
Fever Recognition
Recognizing fever is crucial for early illness detection and response. A fever typically sets in when the body temperature surpasses 100.4掳F (38掳C), indicating the body's response to infection.
  • Measured in Fahrenheit, a fever starts around 100.4掳F.
  • A temperature of 104掳F, like 40.2掳C when converted, is an indicator of a significant fever.
Understanding how different temperatures are classified and how to convert them is essential for managing health effectively.

Recognizing the signs of fever allows for timely medical consultation and appropriate measures to be taken to address any underlying health issues.

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Most popular questions from this chapter

\(\cdot\) Evaporative cooling. The evaporation of sweat is an important mechanism for temperature control in some warmblooded 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 capacity of a typical human body is 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 \(\mathrm{cm}^{3} .\)

\(\cdot\) Heat loss during breathing. In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what is the amount of heat needed to warm to internal body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat capacity of 1.3 \(\mathrm{g}\) is 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) and that 1.0 \(\mathrm{L}\) of air has a mass of 1.3 \(\mathrm{g}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

\(\bullet\) A \(25,000\) -kg subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\)

Jogging in the heat of the day. You have probably seen people jogging in extremely hot weather and wondered "Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 \(\mathrm{kg}\) and surface area 1.85 \(\mathrm{m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{W}, 80 \%\) of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C} .\) (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\) (104 F)? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) . (e) How many 750 \(\mathrm{mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 \(\mathrm{kg}\) .

. 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.
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