Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 44

A family/consumer science class is given an assignment in candy-making that requires a sugar mixture to be brought to a "soft-ball" stage \(\left(234-240^{\circ} \mathrm{F}\right)\). A student borrows a thermometer having a range from \(-10^{\circ} \mathrm{C}\) to \(110^{\circ} \mathrm{C}\) from the chemistry laboratory to do this assignment. Will this thermometer serve the purpose? Explain.

Short Answer

Expert verified
No, the thermometer having a range from \(-10^{\circ} C\) to \(110^{\circ} C\) will not serve the purpose as it does not cover the 'soft-ball' stage temperature range that is \(112.22 - 115.56^{\circ} C\).

Step by step solution

01

Understand the temperature conversion formula

The formula to convert temperature from Fahrenheit to Celsius is given by \(C = (F - 32) \times \frac{5}{9}\). This formula will be used to convert the 'soft-ball' stage temperature range from Fahrenheit to Celsius.
02

Convert the temperature range to Celsius

Apply the conversion formula to the 'soft-ball' stage temperature range (\(234 - 240^{\circ} F\)). The lower limit in Celsius is \((234-32) \times \frac{5}{9} \approx 112.22^{\circ}C\), and the upper limit is \((240-32) \times \frac{5}{9} \approx 115.56^{\circ}C\). So, the temperature range in Celsius is \(112.22 - 115.56^{\circ} C\).
03

Compare the converted temperature range with the thermometer range

The thermometer can measure the temperature from \(-10^{\circ} C\) to \(110^{\circ} C\). Comparing this with the converted 'soft-ball' stage temperature range, \(112.22 - 115.56^{\circ} C\), it is clear that the thermometer does not cover the 'soft-ball' stage temperature range.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Water, a compound, is a substance. Is there any circumstance under which a sample of pure water can exist as a heterogeneous mixture? Explain.

Perform the following calculations; express each answer in exponential form and with the appropriate number of significant figures. (a) \(0.406 \times 0.0023=\) (b) \(0.1357 \times 16.80 \times 0.096=\) (c) \(0.458+0.12-0.037=\) (d) \(32.18+0.055-1.652=\)

Perform the following calculations and retain the appropriate number of significant figures in each result. (a) \(\left(38.4 \times 10^{-3}\right) \times\left(6.36 \times 10^{5}\right)=\) (b) \(\frac{\left(1.45 \times 10^{2}\right) \times\left(8.76 \times 10^{-4}\right)}{\left(9.2 \times 10^{-3}\right)^{2}}=\) (c) \(24.6+18.35-2.98=\) (d) \(\left(1.646 \times 10^{3}\right)-\left(2.18 \times 10^{2}\right)+\left[\left(1.36 \times 10^{4}\right)\right.\) [Hint: The significant figure rule for the extraction of a root is the same as for multiplication.] \(\left.\times\left(5.17 \times 10^{-2}\right)\right]=\) (e) \(\frac{-7.29 \times 10^{-4}+\sqrt{\left(7.29 \times 10^{-4}\right)^{2}+4(1.00)\left(2.7 \times 10^{-5}\right)}}{2 \times(1.00)}\)

Perform the following conversions. (a) \(1.55 \mathrm{kg}=\)________\(\mathrm{g}\) (b) \(642 \mathrm{g}=\)________\(\mathrm{kg}\) (c) \(2896 \mathrm{mm}=\)________\(\mathrm{cm}\) (d) \(0.086 \mathrm{cm}=\)________\(\mathrm{mm}\)

What type of change-physical or chemical-is necessary to separate the following? [Hint: Refer to a listing of the elements.] (a) sugar from a sand/ sugar mixture (b) iron from iron oxide (rust) (c) pure water from seawater (d) water from a slurry of sand in water
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

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.

Sign-up for free