Calorimetry, Specific Heat, and Calculations - AP Chemistry

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Question

Which of the following is the correct molar specific heat of water used when making calculations involving a calorimeter?

Answer

4.184 J/gK is the cited value for the specific heat of water and should be memorized. This is used during calorimeter calculations, specifically when using the equation q= mc delta(T).

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Question

In which instance would a bomb calorimeter be more useful than a coffee-cup calorimeter?

Answer

Bomb calorimeters are most useful when dealing with a gas, because they can operate well at high pressures. Coffee-cup calorimeters are not useful when water begins to boil, producing vapor.

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Question

A 50g sample of a metal was heated to then quickly transferred to an insulated container containing 50g of at . The final temperature of the was .

Which of the following can be concluded?

Answer

When the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water. This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water. However we notice that the water increases by only 5oC and the metal decreases by 65oC. This is becasue of the difference of the specific heats of these substances. The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water.

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Question

How much heat does it take to heat 100g ice at 0C to boiling point?

Cice= 2.1 J/goC

Cwater= 4.2 J/goC

ΔHvap= 2260 J/g

ΔHfus=334 J/g

Answer

You need heat for the phase change, using the enthalpy of fusion (100g334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water (100g100C*4.2 J/goC = 42000 J). Totalling 75400 J (75.4 kJ)

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Question

The specific heat capacity of an unknown liquid is $0.32\frac{J}{kg\cdot K}$ . The density of the liquid is $0.0321 \frac{g}{mL}$ If a chemist applies 243 J of heat to 300 mL of this liquid starting at $27.1^{\circ}C$ , what is the final temperature?

Answer

First we will determine the mass of the liquid:

$300\hspace{1 mm}mL\times\frac{0.0321\hspace{1 mm}g}{1\hspace{1 mm}mL}=9.63\hspace{1 mm}g$

Now we will examine the relationship between heat and specific heat capacity:

$Q=cm\Delta T$

Where $Q$ is heat in Joules, $c$ is the specific heat capacity, $m$ is the mass and $\Delta T$ is the change in temperature. We can rearrange this

$\Delta T=\frac{Q}{cm}$

$\Delta T=\frac{243\hspace{1 mm}J}{0.32J\cdot kg^{-1}\cdot K^{-1}0.00963kg}=78855\hspace{1 mm}K$

If we begin at $27.1^{\circ}C$ , we will end at $78882^{\circ}C$

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Question

The following is a list of specific heat capacities for a few metals.

$\small C_{copper} = 0.385$

$\small C_{iron} = 0.444$

$\small C_{silver} = 0.240$

$\small C_{aluminum} = 0.900$

A 50g sample of an unknown metal is heated with 800 joules. If the temperature of the metal increases by 41.6oC, what is the identity of the unknown metal?

Answer

We need to find the specific heat of the unknown sample of metal in order to locate it on the list. We can do this by using the equation that allows us to determine the specific heat capacity of an element.

$\small q = mC\Delta T$

Since we know the change in temperature, we can simply plug in the values and solve for the value of $\small C$ .

$\small 800J = (50g)C(41.6^oC)$

$\small C = 0.385$

Going back to the list, we see that this is the specific heat capacity for copper, so we confirm that the unknown metal is copper.

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Question

A 20g sample of iron at a temperature of is placed into a container of water. There are 300 milliliters of water in the container at a temperature of .

$\small C_{iron} = 0.444\frac{J}{g^\circ C}$

$\small C_{water} = 4.184\frac{J}{g^\circ C}$

$\small \rho_{water} = 1\frac{g}{mL}$

What is the final temperature of the water?

Answer

There are two things to note before solving for the final temperature.

1. The density of water allows us to say that 300 milliliters of water is the same thing as 300 grams of water.

$300mL*\frac{1g}{1mL}=300g$

2. Since the heat from the iron is being transferred to the water, we can say that the heat transfer is equal between both compounds. Since the heat is conserved in the system, we can set the two equations equal to one another.

$q=mC\Delta T$

$q_i=q_w\rightarrow m_iC_i\Delta T=m_wC_w\Delta T$

$\small m_{i}C_{i}(T_{initial}-T_{final}) = m_{w}C_{w}(T_{final}-T_{initial})$

Notice how the change in temperature for iron has been flipped in order to avoid a negative number.

$\small (20g)(0.444\frac{J}{g^{\circ}C})(120^{\circ}C-T_{f}) = (300g)(4.184\frac{J}{g^{\circ}C})(T_{f}-30^{\circ}C)$

$\small T_{f} = 30.63^{\circ}C$

Because water has a much higher heat capacity compared to iron, the temperature of the water is not changed significantly.

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Question

$\small C_{ice} = 0.5 \frac{cal}{g^{\circ}C}$

$\small C_{water} = 1.0\frac{cal}{g^{\circ}C}$

$\small \Delta H_{fusion} = 80\frac{cal}{g}$

How much energy is needed to raise the temperature of five grams of ice from $\small -10^oC$ to $\small 35^oC$ ?

Answer

This question involves the total energy needed for three different processes: the temperature raise from $\small -10^oC$ to $\small 0^oC$ , the melting of the ice, and the temperature raise from $\small 0^oC$ to $\small 35^oC$ . For the first and third transitions we will use the equation $\small q=mC\Delta T$ . For the melting of ice, we will use the equation $q=m\Delta H_{fusion}$ .

1. $\small \small q =m_{ice}C_{ice}\Delta T = (5g)(0.5\frac{cal}{g^{\circ}C})(10^{\circ}C) = 25 cal$

2. $\small q = m\Delta H_{fusion} = (5g)(80\frac{cal}{g}) = 400 cal$

3. $\small q = m_{water}C_{water}\Delta T = (5g)(1.0\frac{cal}{g^{\circ}C})(35^{\circ}C) = 175 cal$

Finally, we will need to sum the energy required for each step to find the total energy.

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Question

You want to prepare a cup of tea. To do so, you pour of tap water at $20^{0}C$ in a cup that does not absorb microwave radiation and heat it in a microwave oven at of power. If you assume a density of $1.00, \frac{g}{mL}$ for the water and know that its specific heat capacity is $4.184, \frac{J}{g^{o}C}$ , what time do you need to set in the microwave oven to heat the water to $60^{o}C$ ?

Answer

$Q=m\cdot C\cdot\Delta T$

Since the density of water is $1.00, \frac{g}{mL}$ , the mass of is . Plug in known values to the equation and solve.

$100g\cdot\frac{4.184J}{g^{\text o}C}\cdot \left ( 60 - 20 \right )^{\text o}C=16736J$

Use the formula below to find the time needed to heat up the sample of water in the microwave:

$P=\frac{\Delta W}{\Delta t}$

$\Delta t=\frac{16736J}{700W}=23.9s$

Our answer must contain three significant figures.

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Question

How much heat is needed to raise grams of aluminum by $20^{o}:C$ ?

$C_{aluminum}=0.900: \frac{J}{g\cdot^{o}\textrm{C}}$ .

Answer

To find the amount of heat needed to change the temperature of a given material by a certain amount, we'll need to use the equation for specific heat. The specific heat capacity of a compound represents the amount of energy necessary to raise gram of that substance by $1: ^{o}C$ .

$q=mC\Delta T$

$q=(5: grams)(0.900: \frac{J}{g\cdot ^{o}C})(20: ^{o}C)=90:J$

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Question

Which of the following is the correct molar specific heat of water used when making calculations involving a calorimeter?

Answer

4.184 J/gK is the cited value for the specific heat of water and should be memorized. This is used during calorimeter calculations, specifically when using the equation q= mc delta(T).

Compare your answer with the correct one above

Question

In which instance would a bomb calorimeter be more useful than a coffee-cup calorimeter?

Answer

Bomb calorimeters are most useful when dealing with a gas, because they can operate well at high pressures. Coffee-cup calorimeters are not useful when water begins to boil, producing vapor.

Compare your answer with the correct one above

Question

A 50g sample of a metal was heated to then quickly transferred to an insulated container containing 50g of at . The final temperature of the was .

Which of the following can be concluded?

Answer

When the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water. This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water. However we notice that the water increases by only 5oC and the metal decreases by 65oC. This is becasue of the difference of the specific heats of these substances. The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water.

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Question

How much heat does it take to heat 100g ice at 0C to boiling point?

Cice= 2.1 J/goC

Cwater= 4.2 J/goC

ΔHvap= 2260 J/g

ΔHfus=334 J/g

Answer

You need heat for the phase change, using the enthalpy of fusion (100g334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water (100g100C*4.2 J/goC = 42000 J). Totalling 75400 J (75.4 kJ)

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Question

The specific heat capacity of an unknown liquid is $0.32\frac{J}{kg\cdot K}$ . The density of the liquid is $0.0321 \frac{g}{mL}$ If a chemist applies 243 J of heat to 300 mL of this liquid starting at $27.1^{\circ}C$ , what is the final temperature?

Answer

First we will determine the mass of the liquid:

$300\hspace{1 mm}mL\times\frac{0.0321\hspace{1 mm}g}{1\hspace{1 mm}mL}=9.63\hspace{1 mm}g$

Now we will examine the relationship between heat and specific heat capacity:

$Q=cm\Delta T$

Where $Q$ is heat in Joules, $c$ is the specific heat capacity, $m$ is the mass and $\Delta T$ is the change in temperature. We can rearrange this

$\Delta T=\frac{Q}{cm}$

$\Delta T=\frac{243\hspace{1 mm}J}{0.32J\cdot kg^{-1}\cdot K^{-1}0.00963kg}=78855\hspace{1 mm}K$

If we begin at $27.1^{\circ}C$ , we will end at $78882^{\circ}C$

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Question

The following is a list of specific heat capacities for a few metals.

$\small C_{copper} = 0.385$

$\small C_{iron} = 0.444$

$\small C_{silver} = 0.240$

$\small C_{aluminum} = 0.900$

A 50g sample of an unknown metal is heated with 800 joules. If the temperature of the metal increases by 41.6oC, what is the identity of the unknown metal?

Answer

We need to find the specific heat of the unknown sample of metal in order to locate it on the list. We can do this by using the equation that allows us to determine the specific heat capacity of an element.

$\small q = mC\Delta T$

Since we know the change in temperature, we can simply plug in the values and solve for the value of $\small C$ .

$\small 800J = (50g)C(41.6^oC)$

$\small C = 0.385$

Going back to the list, we see that this is the specific heat capacity for copper, so we confirm that the unknown metal is copper.

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Question

A 20g sample of iron at a temperature of is placed into a container of water. There are 300 milliliters of water in the container at a temperature of .

$\small C_{iron} = 0.444\frac{J}{g^\circ C}$

$\small C_{water} = 4.184\frac{J}{g^\circ C}$

$\small \rho_{water} = 1\frac{g}{mL}$

What is the final temperature of the water?

Answer

There are two things to note before solving for the final temperature.

1. The density of water allows us to say that 300 milliliters of water is the same thing as 300 grams of water.

$300mL*\frac{1g}{1mL}=300g$

2. Since the heat from the iron is being transferred to the water, we can say that the heat transfer is equal between both compounds. Since the heat is conserved in the system, we can set the two equations equal to one another.

$q=mC\Delta T$

$q_i=q_w\rightarrow m_iC_i\Delta T=m_wC_w\Delta T$

$\small m_{i}C_{i}(T_{initial}-T_{final}) = m_{w}C_{w}(T_{final}-T_{initial})$

Notice how the change in temperature for iron has been flipped in order to avoid a negative number.

$\small (20g)(0.444\frac{J}{g^{\circ}C})(120^{\circ}C-T_{f}) = (300g)(4.184\frac{J}{g^{\circ}C})(T_{f}-30^{\circ}C)$

$\small T_{f} = 30.63^{\circ}C$

Because water has a much higher heat capacity compared to iron, the temperature of the water is not changed significantly.

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Question

$\small C_{ice} = 0.5 \frac{cal}{g^{\circ}C}$

$\small C_{water} = 1.0\frac{cal}{g^{\circ}C}$

$\small \Delta H_{fusion} = 80\frac{cal}{g}$

How much energy is needed to raise the temperature of five grams of ice from $\small -10^oC$ to $\small 35^oC$ ?

Answer

This question involves the total energy needed for three different processes: the temperature raise from $\small -10^oC$ to $\small 0^oC$ , the melting of the ice, and the temperature raise from $\small 0^oC$ to $\small 35^oC$ . For the first and third transitions we will use the equation $\small q=mC\Delta T$ . For the melting of ice, we will use the equation $q=m\Delta H_{fusion}$ .

1. $\small \small q =m_{ice}C_{ice}\Delta T = (5g)(0.5\frac{cal}{g^{\circ}C})(10^{\circ}C) = 25 cal$

2. $\small q = m\Delta H_{fusion} = (5g)(80\frac{cal}{g}) = 400 cal$

3. $\small q = m_{water}C_{water}\Delta T = (5g)(1.0\frac{cal}{g^{\circ}C})(35^{\circ}C) = 175 cal$

Finally, we will need to sum the energy required for each step to find the total energy.

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Question

You want to prepare a cup of tea. To do so, you pour of tap water at $20^{0}C$ in a cup that does not absorb microwave radiation and heat it in a microwave oven at of power. If you assume a density of $1.00, \frac{g}{mL}$ for the water and know that its specific heat capacity is $4.184, \frac{J}{g^{o}C}$ , what time do you need to set in the microwave oven to heat the water to $60^{o}C$ ?

Answer

$Q=m\cdot C\cdot\Delta T$

Since the density of water is $1.00, \frac{g}{mL}$ , the mass of is . Plug in known values to the equation and solve.

$100g\cdot\frac{4.184J}{g^{\text o}C}\cdot \left ( 60 - 20 \right )^{\text o}C=16736J$

Use the formula below to find the time needed to heat up the sample of water in the microwave:

$P=\frac{\Delta W}{\Delta t}$

$\Delta t=\frac{16736J}{700W}=23.9s$

Our answer must contain three significant figures.

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Question

How much heat is needed to raise grams of aluminum by $20^{o}:C$ ?

$C_{aluminum}=0.900: \frac{J}{g\cdot^{o}\textrm{C}}$ .

Answer

To find the amount of heat needed to change the temperature of a given material by a certain amount, we'll need to use the equation for specific heat. The specific heat capacity of a compound represents the amount of energy necessary to raise gram of that substance by $1: ^{o}C$ .

$q=mC\Delta T$

$q=(5: grams)(0.900: \frac{J}{g\cdot ^{o}C})(20: ^{o}C)=90:J$

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