/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 While vacationing in Italy, you ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 3

While vacationing in Italy, you see on local TV one summer morning that temperature will rise from the current \(18^{\circ} \mathrm{C}\) to a high of \(39^{\circ} \mathrm{C}\) . What is the corresponding increase in the Fahrenheit temperature?

Short Answer

Expert verified
The increase in Fahrenheit temperature is \(37.8^{\circ} \mathrm{F}\).

Step by step solution

01

Understand the Celsius to Fahrenheit Formula

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

Convert Initial Celsius Temperature to Fahrenheit

First, convert the initial temperature of \(18^{\circ} \mathrm{C}\) to Fahrenheit:\[ F_1 = \frac{9}{5} \times 18 + 32 \]Calculating gives:\[ F_1 = 32.4 + 32 = 64.4^{\circ} \mathrm{F} \]
03

Convert Final Celsius Temperature to Fahrenheit

Next, convert the high temperature of \(39^{\circ} \mathrm{C}\) to Fahrenheit:\[ F_2 = \frac{9}{5} \times 39 + 32 \]Calculating gives:\[ F_2 = 70.2 + 32 = 102.2^{\circ} \mathrm{F} \]
04

Calculate the Increase in Fahrenheit

Subtract the initial Fahrenheit temperature from the final Fahrenheit temperature:\[ \Delta F = F_2 - F_1 = 102.2 - 64.4 \]This calculation results in:\[ \Delta F = 37.8^{\circ} \mathrm{F} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Celsius to Fahrenheit Formula
When you want to convert a temperature from Celsius to Fahrenheit, there's a straightforward formula that you can use: \( F = \frac{9}{5}C + 32 \). This means you'll need to multiply the Celsius temperature by \( \frac{9}{5} \) and then add 32 to the result.
For example, if you have a temperature of \(10^{\circ} \mathrm{C}\), you would calculate \( F = \frac{9}{5} \times 10 + 32 = 18 + 32 = 50^{\circ} \mathrm{F} \).
It's a good idea to memorize this formula for converting Celsius to Fahrenheit, as it can be very handy, especially when traveling or dealing with weather forecasts in different regions.
Fahrenheit Temperature Calculation
Converting temperatures from Celsius to Fahrenheit involves calculating each temperature separately using the formula mentioned earlier.
Let's look at the given example: You start with \(18^{\circ} \mathrm{C}\), converting to Fahrenheit using \( F_1 = \frac{9}{5} \times 18 + 32 \).
Breaking it down:
  • Multiply 18 by \(\frac{9}{5} \) to get 32.4.
  • Add 32 to 32.4 to get 64.4. So, \(18^{\circ} \mathrm{C}\) is the same as \(64.4^{\circ} \mathrm{F}\).
Next, for \(39^{\circ} \mathrm{C}\), use the formula: \( F_2 = \frac{9}{5} \times 39 + 32 \).
Breaking down this calculation:
  • Multiply 39 by \(\frac{9}{5} \) to get 70.2.
  • Add 32 to 70.2 to reach 102.2. Thus, \(39^{\circ} \mathrm{C}\) is \(102.2^{\circ} \mathrm{F}\).
These simple steps help convert temperatures accurately for any situation.
Temperature Increase Calculation
Once you've converted both Celsius temperatures into Fahrenheit, calculating the increase in temperature becomes easy.
You simply subtract the initial Fahrenheit temperature from the final Fahrenheit temperature. In the exercise, you started with \(64.4^{\circ} \mathrm{F}\) and concluded with \(102.2^{\circ} \mathrm{F}\).
To find the increase, compute:
  • \(\Delta F = F_2 - F_1 = 102.2 - 64.4 = 37.8^{\circ} \mathrm{F}\).
This calculation reveals that the Fahrenheit temperature will increase by \(37.8^{\circ} \mathrm{F}\).
Understanding this process is essential for grasping how temperature changes are measured across different temperature scales.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C} .\) By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C} ?\)

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.

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 amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat of air is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that 1.0 \(\mathrm{L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{kg}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

On a cool \(\left(4,0^{\circ} \mathrm{C}\right)\) Saturfay moming, a pilot fills the fuel tanks of her Pitts \(S-2 C\) (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again \(4.0^{\circ} \mathrm{C}\) , she checks the fuel level and finds only 103.4 \(\mathrm{L}\) of gasoline in the tanks. She realizes that it was hot on Saturday afternoon, and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in "C) reached by the fuel and the tank on Saturday aftemoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10^{-4} \mathrm{K}^{-1}\) , and the tank is made of aluminum. (b) In order to have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

"The Ship of the Desert" Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 \(\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Modern Physics

Read Explanation

Famous Physicists

Read Explanation

Medical Physics

Read Explanation

Thermodynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.