/*! 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} Q2E Heat stroke .If the body’s tem... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Heat stroke .If the body’s temperature is above 105°F for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

Short Answer

Expert verified

The temperature in Celsius scale is 40.6°C

The temperature in Kelvin scale is 313.8K

Given :

The temperature in Fahrenheit scale is 105°F

Step by step solution

01

Step 1:

To covert the temperature on Fahrenheit scale into Celsius scale.

Temperature in Fahrenheit scale can be converted in temperature in Celsius scale by formula :

Tc=59TF-32

Where is the TFtemperature in Fahrenheit scale and TCtemperature in Celsius scale.

To covert the temperature on Celsius scale to Kelvin scale

Temperature in Celsius scale can be converted in temperature in Kelvin scale by formula :

TK=Tc+273.15

Where is the Tctemperature in Celsius scale TKtemperature is Kelvin scale.

02

Calculation of temperature in Celsius scale and Kelvin scale

For conversion of temperature from Fahrenheit scale to Celsius scale

Using formula

Tc=59TF-32

Now putting the values of constants in above equation

Tc=59105-32Tc=59×73Tc=40.6°C

Thus, the temperature in Celsius scale is 40.6°C.

For conversion of temperature from Celsius scale to Kelvin scale

Using formula

TK=Tc+273.15

Now, putting the values of constant in above equation

TK=40.6+273.15TK=313.8K

This,the temperature in Kelvin scale is 313.8K

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 91Ó°ÊÓ!

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

Some critics of biological evolution claim that it violates the second law of thermodynamics, since evolution involves simple life forms developing into more complex and more highly ordered organisms. Explain why this is not a valid argument against evolution.

A gas storage tank has a small leak. The pressure in the tank drops more quickly if the gas is hydrogen or helium than if it is oxygen. Why?

It is well known that a potato bakes faster if a large nail is suck through it .Why? Does an Aluminium nail work better than a steel one? Why or why not? There is also a gadget on the market to hasten the roasting meat; it consists of a hollow metal tube containing a wick and some water. This is claimed to work much better than a solid metal rod. How does it work?

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at 19.0°C. What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen (77.3 K)?

The efficiency of heat engines is high when the temperature difference between the hot and cold reservoirs is large. Refrigerators, on the other hand, work better when the temperature difference is small. Thinking of the mechanical refrigeration cycle shown in Fig. 20.9, explain in physical terms why it takes less work to remove heat from the working substance if the two reservoirs (the inside of the refrigerator and the outside air) are at nearly the same temperature, than if the outside air is much warmer than the interior of the refrigerator.

See all solutions

Recommended explanations on Physics Textbooks

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.