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High School Physics : Understanding Heat and Temperature

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Example Questions

Example Question #1 : Understanding Heat And Temperature

The temperature of an ideal gas is raised from $18.5^{\circ}C$ to $74.33^{\circ}C$ . If the volume remains constant, what was its initial pressure if the final pressure is 81Pa ?

Possible Answers:

10.08Pa

25.17Pa

20.16Pa

4522.23Pa

325.44Pa

Correct answer:

20.16Pa

Explanation:

For this problem, use Gay-Lussac's law to set up a direct proportion between pressure and temperature. Note that this law only applies when volume is constant.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Plug in our given values and solve for the initial pressure.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{18.5^{\circ}C}=\frac{81Pa}{74.33^{\circ}C}$

$\frac{P_1}{18.5^{\circ}C}=1.09\frac{Pa}{^\circ C}$

$P_1=1.09\frac{Pa}{^\circ C}* 18.5^\circC$

P_1=20.16Pa

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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.

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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 $22^\circ C$ .

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

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

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

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

$-45.29^\circ C=-T_i$

$45.29^\circ C=T_i$

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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 $80^\circ C$ .

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

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

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

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

$52.67^\circ C=T_f$

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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.

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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

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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.

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

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

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

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

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

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

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

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

$T_f=78.43^\circ C$

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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.

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

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

$T_f=95.85^\circ C$

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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.

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

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

$T_f=70.23^\circ C$

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High School Physics : Understanding Heat and Temperature

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 #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

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