Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Heat

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

Example Question #1 : Understanding Heat And Temperature

0.75kg of soup at $80^\circ C$ cools down to $20^\circ C$ after 2300s . If the specific heat of the soup is $\small 3.67\frac{kJ}{kg^\circ C}$ , how much energy does the soup release into the room?

Possible Answers:

1651.5kJ

165.15kJ

0.72kJ

27.53kJ

275.25kJ

Correct answer:

165.15kJ

Explanation:

The formula for heat energy is:

$Q=mc\Delta t$

We are given the initial and final temperatures, mass, and specific heat. Using these values, we can find the heat released. Note that the time is irrelevant to this calculation.

$Q=(0.75kg)(3.67\frac{kJ}{kg^\circ C})(20^\circ C -80^\circ C)$

Q=-165.15kJ

That means that the soup "lost" 165.15kJ of energy. This is the amount that it released into the room. The value is negative for the soup, the source of the heat, but positive for the room, which receives it.

Report an Error

Example Question #3 : Understanding Heat And Temperature

0.75kg of soup cools down to $22^\circ C$ after 2800s . If the specific heat of the soup is $\small 3.67\frac{kJ}{kg^\circ C}$ , and it released 64.11kJ of energy into the room, what was the initial temperature of the soup?

Possible Answers:

$45.29^\circ C$

$1.29^\circ C$

$-1.29^\circ C$

$0.03^\circ C$

$24.58^\circ C$

Correct answer:

$45.29^\circ C$

Explanation:

The formula for heat energy is:

$Q=mc\Delta t$

We are given the final temperature, mass, specific heat, and heat released. Using these values, we can find the initial temperature. Note that the time is irrelevant to this calculation. Since heat is released from the soup, the net change in the soup's energy is negative. Since the soup is cooling, we expect our answer to be greater than $22^\circ C$ .

$-64.11kJ=(0.75kg)(3.67\frac{kJ}{kg^\circ C})(22^\circ C- T_i)$

$-64.11kJ=2.7525\frac{kJ}{^\circ C}*(22^\circ C - T_i)$

$\frac{-64.11kJ}{2.7525\frac{kJ}{^\circ C}}=(22^\circ C -T_i )$

$-23.29^\circ C=22^\circ C - T_i$

$-45.29^\circ C=-T_i$

$45.29^\circ C=T_i$

Report an Error

Example Question #4 : Understanding Heat And Temperature

0.75kg of soup at $80^\circ C$ cools down after 2300s . If the specific heat of the soup is $\small 3.67\frac{kJ}{kg^\circ C}$ , and it released 75.22kJ of energy into the room, what is the final temperature of the soup?

Possible Answers:

$43.67^\circ C$

$0.023^\circ C$

$107.33^\circ C$

$52.67^\circ C$

$27.33^\circ C$

Correct answer:

$52.67^\circ C$

Explanation:

The formula for heat energy is:

$Q=mc\Delta t$

We are given the initial temperature, mass, specific heat, and heat released. Using these values, we can find the final temperature. Note that the time is irrelevant to this calculation. Since heat is released from the soup, the net change in the soup's energy is negative. Since the soup is cooling, we expect our answer to be less than $80^\circ C$ .

$-75.22kJ=(0.75kg)(3.67\frac{kJ}{kg^\circ C})(T_f -80^\circ C)$

$-75.22kJ=(2.7525\frac{kJ}{^\circ C})(T_f-80^\circ C)$

$\frac{-75.22kJ}{2.7525\frac{kJ}{^\circ C}}=(T_f-80^\circ C)$

$-27.33^\circ C=Tf-80^\circ C$

$52.67^\circ C=T_f$

Report an Error

Example Question #21 : Thermodynamics

An ice cube at $0^\circ C$ melts. As it melts, constant temperature readings are taken and the sample maintains the temperature of $0^\circ$ throughout the melting process. Which statement best describes the energy of the system?

Possible Answers:

We would need to know the mass of the ice cube to draw a conclusion

The form of water has no bearing on the system energy

Energy increases as the sample moves from a solid to a liquid

Energy is being used to convert the ice to water

Energy of the system remains the same

Correct answer:

Energy is being used to convert the ice to water

Explanation:

When an object changes phase, it requires energy called "latent heat." In this case, even though the temperature is remaining constant, the energy inside of the ice cube is decreasing as it expends energy to melt.

Report an Error

Example Question #1 : Understanding Heat And Temperature

A 280K silver spoon is placed in a 366K cup of tea. If the spoon has a mass of 32g and the tea has 90g of mass, what is the final temperature of the spoon?

$c_{tea}=1.00\frac{cal}{g*K}$

$c_{silver}= 0.0558\frac{cal}{g*K}$

Possible Answers:

288.32K

353.43K

323K

364.33K

301.21K

Correct answer:

364.33K

Explanation:

The equation for two items reaching a thermal equilibrium is given by describing a heat transfer. The heat removed from one object is equal to the heat added to the other.

K_1=-K_2

$K=mc\Delta T$

$m_1c_1\Delta T_1=-m_2c_2\Delta T_2$

We are given the specific heat values of each substance, as well as their masses. We also know the initial temperature of each substance. Use these terms in the equation to solve for the final temperature. Remember that the final temperature will be the same for each substance, since they will be in thermodynamic equilibrium.

$(32g)(0.0558\frac{cal}{g*K})(T_f-280K)=-(90g)(1.00\frac{cal}{g*K})(T_f-366K)$

$1.7856\frac{cal}{K}*(T_f-280K)=-90\frac{cal}{g*K}*(T_f-366K)$

$(1.7856\frac{cal}{K}*T_f)-499.968cal=(-90\frac{cal}{K}*T_f)+32940cal$

$1.7856\frac{cal}{K}*T_f=(-90\frac{cal}{K}*T_f)+33439.968cal$

$91.7856\frac{cal}{K}*T_f=33439.968cal$

$T_f=\frac{33439.968cal}{91.7856\frac{cal}{K}}$

T_f=364.33K

Report an Error

Example Question #1 : Understanding Heat And Temperature

A sample of 236g of water at $100^\circ C$ is placed in a 350g ceramic mug, which is at $23^\circ C$ . What is the final temperature of the system?

$c_{ceramic}=1100\frac{J}{kg*^\circ C}$

$c_{water}=4.2\frac{J}{g*^\circ C}$

Possible Answers:

$123^\circ C$

$78.43^\circ C$

$84^\circ C$

$91^\circ C$

$108^\circ C$

Correct answer:

$78.43^\circ C$

Explanation:

For this question, we must recognize that the system going to end up in equilibrium. That means that:

$m_1c_1\Delta T_1=-m_2c_2\Delta T_2$

We are given the initial temperatures, masses, and specific heats of both the water and the ceramic. This will allow us to solve for the final temperature of the system; this value will be equal for both components. Notice that the specific heat given to us in the problem for the ceramic is in terms of kilograms, not grams. Convert to grams.

$(350g)(1.1\frac{J}{g*^\circ C})(T_f-23^\circ C)=-(236g)(4.2\frac{J}{g*^\circ C})(T_f-100^\circ C)$

$(385\frac{J}{^\circ C})(T_f-23^\circ C)=-(991.2\frac{J}{^\circ C})(T_f-100^\circ C)$

$\frac{(385\frac{J}{^\circ C})(T_f-23^\circ C)}{385\frac{J}{^\circ C}}=\frac{(-991.2\frac{J}{^\circ C})(T_f-100^\circ C)}{385\frac{J}{^\circ C}}$

$T_f-23^\circ C=(-2.57)(T_f-100^\circ C)$

$T_f-23^\circ C=-2.57T_f+257^\circ C$

$-23^\circ C=-3.57T_f+257^\circ C$

$-280^\circ C=-3.57T_f$

$\frac{-3.57T_f}{-3.57}=\frac{-280^\circ C}{-3.57}$

$T_f=78.43^\circ C$

Report an Error

Example Question #1 : Understanding Heat And Temperature

A 78g vial of an unknown liquid is $9 8^\circ C$ . Julie adds of the same liquid at $40^\circ C$ to the vial. What is the final temperature?

Possible Answers:

We need to know the freezing/boiling points of the liquid in order to solve.

$94.74^\circ C$

$7 7 . 6 4 ^\circ C$

We need to know the specific heat of the liquid in order to solve.

$95.85^\circ C$

Correct answer:

$95.85^\circ C$

Explanation:

The equation for change in temperature is $m_1c_1\Delta T_1=-m_2c_2\Delta T_2$

Plug in our given values.

$m_1c_1\Delta T_1=-m_2c_2\Delta T_2$

$0.78kg*c*(T_f-98^\circ C)=-0.03kg*c*(T_f -40^\circ C)$

Notice that the specific heats will cancel out.

$0.78kg*(T_f-98^\circ C)=-0.03kg*(T_f -40^\circ C)$

$(0.78kg*T_f)-(0.78kg*98^\circ C)=(-0.03kg*T_f)+(0.03kg*40^\circ C)$

$(0.78kg*T_f)-(76.44kg*^\circ C)=(-0.03kg*T_f)+(1.2kg*^\circ C)$

Combine like terms.

$(0.78kg*T_f)-(-0.03kg*T_f)=(1.2kg*^\circ C)+(76.44kg*^\circ C)$ $0.81kg*T_f=77.64kg*^\circ C$

$T_f=\frac{77.64kg*^\circ C}{0.81kg}{}$

$T_f=95.85^\circ C$

Report an Error

Example Question #1 : Thermodynamics

A disc of copper is dropped into a 12g glass of water. If the copper was at $85^\circ C$ and the water was at $70^\circ C$ , what is the new temperature of the mixture?

$c_{Cu}=0.092\frac{cal}{g^oC}$

$c_{water}=1.00\frac{cal}{g^oC}$

Possible Answers:

70.23^oC

75.22^oC

79.77^oC

69.77^oC

83.76^oC

Correct answer:

70.23^oC

Explanation:

The relationship between mass and temperature, when two masses are mixed together, is:

$m_1c_1\Delta T=-m_2c_2\Delta T$

Using the given values for the mass and specific heat of each compound, we can solve for the final temperature.

$(2g)(0.092\frac{cal}{g*^\circ C})(T_f-85^\circ C)=-(12g)(1\frac{cal}{g*^\circ C})(T_f-70^{\circ} C)$

We need to work to isolate the final temperature.

$(0.184\frac{cal}{^\circ C})(T_f-85^\circ C)=-(12\frac{cal}{^\circ C})(T_f-70^{\circ} C)$

Distribute into the parenthesis using multiplication.

$(0.184\frac{cal}{^\circ C})T_f-15.64cal=(-12\frac{cal}{^\circ C})T_f+840cal$

Combine like terms.

$(12.184\frac{cal}{^oC})T_f=855.64cal\textup{}$

$T_f=\frac{855.64cal}{12.184\frac{cal}{^oC}}$

$T_f=70.23^\circ C$

Report an Error

Example Question #1 : Understanding Effects Of Heat On Volume

An ideal gas is inside of a container with a pressure of 2Pa . If it starts with a volume of and is compressed to 0.5L , what is the new pressure if the temperature remains constant?

Possible Answers:

10Pa

8Pa

4Pa

2Pa

0.5Pa

Correct answer:

4Pa

Explanation:

We will need to use Boyle's Law to solve:

P_1V_1=P_2V_2

Boyle's Law allows us to set up a relationship between the changes in pressure and volume under conditions with constant temperature. Since the equation is a proportion, we do not need to convert any units.

We can use the given values to solve for the new pressure.

(2Pa)(1L)=(P_2)(0.5L)

$\frac{(2Pa)(1L)}{0.5L}=(P_2)$

$\frac{2Pa\cdot L}{0.5L}=P_2$

4Pa=P_2

Report an Error

Example Question #2 : Understanding Effects Of Heat On Volume

An ideal gas is inside of a tube at 60^oC . If the pressure remains constant, but the volume decreases from 5m^3 to 3m^3 , what will be the final temperature in the tube?

Possible Answers:

36^oC

180^oC

18^oC

24^oC

120^oC

Correct answer:

36^oC

Explanation:

For this problem, use Charles's Law:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

In this formula, is the volume and is the temperature. Charles's Law allows us to set up a proportion for changes in volume and temperature, as long as pressure remains constant. Since we are dealing with a proportion, the units for temperature are irrelevant and we do not need to convert to Kelvin.

Using the given values, we should be able to solve for the final temperature.

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$\frac{5m^3}{60^oC}=\frac{3m^3}{T_2}$

Cross multiply.

5m^3* T_2=3m^3* 60^oC

$5m^3* T_2=180(^oC\cdot m^3)$

$T_2=\frac{180(^{\circ}C\cdot m^3)}{5m^3}$

$T_2=36^{\circ}C$

Report an Error

← Previous 1 2 3 Next →

View Tutors

Lazaro
Certified Tutor

Miami Dade College, Bachelor of Engineering, Computer Hardware Engineering.

View Tutors

Dylan
Certified Tutor

Oberlin College, Bachelors, Mathematics.

View Tutors

Alexandra
Certified Tutor

Rice University, Bachelor in Arts, Biochemistry and Molecular Biology. University of North Texas Health Science Center, Maste...

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Dallas Fort Worth, Calculus Tutors in New York City, Biology Tutors in Chicago, LSAT Tutors in Philadelphia, French Tutors in Houston, Reading Tutors in Boston, GRE Tutors in San Diego, GRE Tutors in Phoenix, LSAT Tutors in New York City, GMAT Tutors in Los Angeles

Popular Courses & Classes

GRE Courses & Classes in Phoenix, LSAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Boston, GMAT Courses & Classes in Atlanta, Spanish Courses & Classes in Washington DC, ACT Courses & Classes in New York City, LSAT Courses & Classes in Miami, GRE Courses & Classes in Washington DC, SAT Courses & Classes in Chicago, ACT Courses & Classes in Los Angeles

Popular Test Prep

ACT Test Prep in Miami, SAT Test Prep in Chicago, SAT Test Prep in Miami, ACT Test Prep in New York City, LSAT Test Prep in Los Angeles, GRE Test Prep in Chicago, LSAT Test Prep in Denver, GMAT Test Prep in Atlanta, ACT Test Prep in Seattle, LSAT Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Physics : Heat

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Understanding Heat And Temperature

Example Question #3 : Understanding Heat And Temperature

Example Question #4 : Understanding Heat And Temperature

Example Question #21 : Thermodynamics

Example Question #1 : Understanding Heat And Temperature

Example Question #1 : Understanding Heat And Temperature

Example Question #1 : Understanding Heat And Temperature

Example Question #1 : Thermodynamics

Example Question #1 : Understanding Effects Of Heat On Volume

Example Question #2 : Understanding Effects Of Heat On Volume

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: