Understanding Heat and Temperature - High School Physics

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Question

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 ?

Answer

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.

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

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

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

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

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Question

A disc of copper is dropped into a 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?

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

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

Answer

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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Question

of soup at $80^\circ C$ cools down to $20^\circ C$ after . 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?

Answer

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.

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

That means that the soup "lost" 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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Question

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

Answer

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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Question

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

Answer

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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Question

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?

Answer

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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Question

A silver spoon is placed in a cup of tea. If the spoon has a mass of and the tea has of mass, what is the final temperature of the spoon?

$c_{tea}=1.00\frac{cal}{gK}$

$c_{silver}= 0.0558\frac{cal}{gK}$

Answer

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.

$K=mc\Delta T$

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

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.

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

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

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

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

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

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

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Question

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

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

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

Answer

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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Question

A 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?

Answer

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

Plug in our given values.

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

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

Notice that the specific heats will cancel out.

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

$(0.78kgT_f)-(0.78kg98^\circ C)=(-0.03kgT_f)+(0.03kg40^\circ C)$

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

Combine like terms.

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

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

$T_f=95.85^\circ C$

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