Question 1

Use the following to answer question : 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the solid can be found by

Accepted Answer

The specific heat of a substance is defined as the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. From the graph, we can see that the specific heat of the solid is given by the ratio of the rate at which heat is added (in watts) to the product of the slope of segment A (in degrees Celsius per second) and the mass of the substance (in grams). Therefore, the correct choice is C.

Question 2

Two liquids, A and B, are mixed together. Liquid A has mass m and was initially at temperature 40S1U1P1$\circ$S1S1P0C, and liquid B has mass 2m and was initially at temperature 5S1U1P1$\circ$S1S1P0C. The specific heat of liquid A is 1.5 times that of liquid B.
Calculate the final temperature of the mixture.

Accepted Answer

Using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can set up two equations to represent the heat transferred from each liquid and set them equal to each other.

For liquid A: Q = (m)(1.5c)(Tf-40) 
For liquid B: Q = (2m)(c)(Tf-5)

Since the heat transferred between the two liquids is equal, we can set the equations equal to each other and solve for Tf:

(m)(1.5c)(Tf-40) = (2m)(c)(Tf-5)

Simplifying and solving for Tf:

1.5(Tf-40) = 2(Tf-5) 
1.5Tf - 60 = 2Tf - 10 
50 = 0.5Tf 
Tf = 100

So the final temperature of the mixture is 100F1F1F1S1F1F1F10C, which when converted to Celsius is approximately 20.0F1F1F1S1F1F1F10C. Therefore, the best choice is D.

Question 3

A 250-g piece of lead is heated to 100ºC and is then placed in a 400-g copper container holding 500 g of water. The specific heat of copper is c = 0.386 kJ/kg · K. The container and the water had an initial temperature of 18.0ºC. When thermal equilibrium is reached, the final temperature of the system is 19.15ºC. If no heat has been lost from the system, what is the specific heat of the lead? (the specific heat of water is 4.180 kJ/kg · K)

Accepted Answer

We can use the principle of conservation of energy to solve this problem. The heat lost by the lead is equal to the heat gained by the water and the copper container:

Qlost = Qgained

The heat lost by the lead can be calculated using the formula:

Qlost = mlead x clead x deltaTlead

where mlead is the mass of lead, clead is the specific heat of lead, and deltaTlead is the change in temperature of lead. Since the lead is heated from 18.0ºC to 100ºC, deltaTlead = 82.0ºC.

The heat gained by the water and copper container can be calculated using the formula:

Qgained = (mwater + mcopper) x c x deltaT

where mwater and mcopper are the masses of water and copper, c is the specific heat of copper, and deltaT is the change in temperature of the water and copper. Since the initial temperature of the water and copper is 18.0ºC and the final temperature is 19.15ºC, deltaT = 1.15ºC.

Substituting these values into the equation Qlost = Qgained and solving for clead, we get:

mlead x clead x deltaTlead = (mwater + mcopper) x c x deltaT

clead = [(mwater + mcopper) x c x deltaT] / (mlead x deltaTlead)

Substituting the given values, we get:

clead = [(0.500 kg + 0.400 kg) x 0.386 kJ/kg · K x 1.15ºC] / (0.250 kg x 82.0ºC) = 0.128 kJ/kg · K

Therefore, the specific heat of lead is 0.128 kJ/kg · K, which is closest to choice B.

Question 4

Body A has twice the mass and three times the specific heat of body B. They are supplied with equal amounts of heat. Body A experiences a temperature change $\Delta$T. What change in temperature is experienced by body B?

Accepted Answer

The answer of Body A has twice the mass and...

Question 5

Use the following to answer question : 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the liquid can be found by

Accepted Answer

The specific heat of the liquid can be found by dividing the rate at which heat is added (Q/t) by the product of the slope of C and the mass of the substance (msC). This is because during this segment, the substance is undergoing a phase change from liquid to vapor, and the heat added is being used to overcome the heat of vaporization (which is proportional to the mass of the substance) rather than to increase the temperature. Therefore, the specific heat cannot be calculated using the length of B or D (which represent segments where the temperature is increasing) or the slope of A or E (which are not relevant to the phase change).

Question 6

Aluminum has a specific heat more than twice that of copper. Identical masses of aluminum and copper, both at 0ºC, are dropped together into a can of hot water. When the system has come to equilibrium,

Accepted Answer

The final temperature reached by both aluminum and copper will be the same as they are in the same system and will reach thermal equilibrium. The specific heat capacity of a material only affects how much heat energy is required to raise the temperature of the material, not the final temperature it will reach when in thermal equilibrium with other objects.

Question 7

To raise the temperature of a 2.0-kg piece of metal from 20º to 100ºC, 61.8 kJ of heat is added. What is the specific heat of this metal?

Accepted Answer

The specific heat (c) can be found using the equation:
Q = mcΔT
where Q is the amount of heat added, m is the mass of the metal, c is the specific heat, and ΔT is the change in temperature.
Substituting the given values and solving for c, we get:
61.8 kJ = (2.0 kg) c (100ºC - 20ºC)
c = 0.39 kJ/kg · K. Therefore, the answer is A.

Question 8

Two liquids, A and B, are mixed together, and the resulting temperature is 22S1U1P1$\circ$S1S1P0C. If liquid A has mass m and was initially at temperature 35S1U1P1$\circ$S1S1P0C, and liquid B has mass 3m and was initially at temperature 11S1U1P1$\circ$S1S1P0C, calculate the ratio of the specific heats of A divided by B.

Accepted Answer

We can use the principle of conservation of energy to solve the problem. The heat lost by liquid A must equal the heat gained by liquid B. We can express this as:
$mC_A(T_f -35) = 3mC_B(T_f-11)$
where $C_A$ and $C_B$ are the specific heats of liquids A and B, and $T_f$ is the final temperature of the mixture.
We also know that the specific heat of a substance is related to its heat capacity per unit mass:
$C = \frac{Q}{m\Delta T}$
where $Q$ is the amount of heat transferred, $m$ is the mass, and $\Delta T$ is the change in temperature. We can rewrite this as:
$C = \frac{1}{m}\left(\frac{Q}{\Delta T}\right)$
Therefore, we can express the ratio of specific heats as:
$\frac{C_A}{C_B} = \frac{1}{m}\left(\frac{Q_A}{\Delta T_A}\right) \div \frac{1}{3m}\left(\frac{Q_B}{\Delta T_B}\right) = \frac{3Q_A\Delta T_B}{Q_B\Delta T_A}$
where $Q_A$ and $Q_B$ are the heats transferred between the liquids, and $\Delta T_A$ and $\Delta T_B$ are their respective temperature changes.
Now we need to find the unknowns $Q_A$ and $Q_B$. We can use the formula for heat transfer by conduction:
$Q = kA\Delta T$
where $k$ is the thermal conductivity of the material, $A$ is the area of contact, and $\Delta T$ is the temperature difference. Assuming that the liquids mix thoroughly, we can assume $A$ is the same for both liquids. Therefore, we can write:
$Q_A = kA\Delta T_A = kA(T_f - 22)$
$Q_B = kA\Delta T_B = kA(22)$
Substituting these into the expression for the ratio of specific heats, we get:
$\frac{C_A}{C_B} = \frac{3(T_f-22)}{22(T_f-35)}$
We can solve for $T_f$ by plugging in the given values and simplifying:
$22(T_f-22) = 3mC_A(T_f-35) = 9mC_B(T_f-11)$
$22T_f - 484 = 9T_f - 99m$
$T_f = \frac{484 + 99m}{13}$
Substituting this back into the expression for the ratio of specific heats, we get:
$\frac{C_A}{C_B} = \frac{3(\frac{484+99m}{13}-22)}{22(\frac{484+99m}{13}-35)} = \frac{165-33m}{418-99m}$
Now we can test each answer choice to see which one produces the correct ratio. Only choice B yields the correct result of $\frac{C_A}{C_B} = 2.5$.

Question 9

The specific heat of a substance is a measure of

Accepted Answer

Specific heat is defined as the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. It is a measure of the amount of energy needed to change the kinetic energies of atoms and/or molecules and/or electrons in a substance by a temperature increment.

Question 10

At liquid ⁴He temperatures (< 4.2 K), the molar heat capacity heat of most insulators goes as C = AT³, where A is a constant and T is in kelvins. The ratio of the heat required to raise the temperature from 2 K to 3K and from 1 K to 2K is

Accepted Answer

The answer of At liquid ⁴He temperatures (< 4.2 K),...

Question 11

A system has a heat capacity of 100 J. This means

Accepted Answer

Heat capacity is a measure of the amount of heat energy required to change the temperature of a system. It does not imply that all of this energy can be converted to work (due to the second law of thermodynamics), but it does mean that heat can be transferred to or from the system, including potentially doing some work or transferring heat to another system if there is a temperature difference.

Question 12

The quantity of heat absorbed by a body is determined from the formula Q = cm(T_f - T_i). A certain metal has a specific heat c = 0.21 cal/g Cº and a mass m = 25.6 g. The initial temperature is T_i = 34.6ºC, and the final temperature T_f = 54.6ºC. The quantity of heat absorbed is

Accepted Answer

The answer of The quantity of heat absorbed by a...

Question 13

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg · Cº) at a temperature of 430ºC is dropped into 48 kg of water. The water is initially at a temperature of 10ºC. With no heat losses to the surroundings, the equilibrium temperature of the iron and water is approximately

Accepted Answer

The equilibrium temperature can be found using the principle of conservation of energy, where the heat lost by the iron is equal to the heat gained by the water. The formula for heat (Q) is Q = mcΔT, where m is mass, c is specific heat, and ΔT is the change in temperature. For iron, Q_iron = m_iron * c_iron * (T_final - T_initial_iron), and for water, Q_water = m_water * c_water * (T_final - T_initial_water). Since Q_iron = -Q_water (heat lost by iron is gained by water), we can set their magnitudes equal and solve for T_final. Given the specific heat of water is approximately 1 kcal/kg·ºC, and using the given values, solving the equation yields a final temperature close to 12ºC, indicating that option A is correct.

Question 14

Use the following to answer question : 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated at constant volume. This process is shown in the graph. The specific heat at constant volume of the vapor can be found by

Accepted Answer

The answer of Use the following to answer question :...

Question 15

For most metals, the heat capacity goes as C = $\gamma$T + AT³ at low temperatures (~ 100 K), whereas insulators only has the T³ term. Both $\gamma$ and A are constants. The two terms in the expression for C suggests

Accepted Answer

The two terms suggest that there are two independent mechanisms for heat to be absorbed or released in metals. The first term (F1F1F1S1F1F1F10T) represents the contribution to heat capacity from lattice vibrations or phonons, while the second term (AT³) represents the contribution from electrons. Insulators do not have the contribution from electrons, so they only have the T³ term. Therefore, the two terms in the expression for C distinguish between the behavior of metals and insulators.

Question 16

A 1.0-kg piece of marble at 100ºC is dropped into 2.5 kg of water at 1.0ºC and the resulting temperature is 7.0ºC. The specific heat of the marble is approximately

Accepted Answer

The answer of A 1.0-kg piece of marble at 100ºC...

Question 17

Glass beads of mass 100 g and specific heat 0.20 cal/g · Cº are heated to 90ºC and placed in a 300-g glass beaker containing 200 g of water at 20ºC. When equilibrium is reached, the temperature is approximately

Accepted Answer

Using the formula Q = mcΔT, where Q is the heat gained/lost, m is the mass, c is the specific heat, and ΔT is the change in temperature, we can find the amount of heat lost by the glass beads and gained by the water:
Q(beads) = (100 g)(0.20 cal/g · Cº)(90ºC - Tf)
Q(water) = (200 g)(1 cal/g · Cº)(Tf - 20ºC)
Since Q(beads) = -Q(water) (heat lost = heat gained), we can set them equal to each other and solve for Tf:
(100 g)(0.20 cal/g · Cº)(90ºC - Tf) = (200 g)(1 cal/g · Cº)(Tf - 20ºC)
1800 - 20Tf = 200Tf - 4000
220Tf = 5800
Tf ≈ 26.4ºC
Therefore, the temperature of the system at equilibrium is approximately 26.4ºC, which is closest to answer choice B (25ºC).

Question 18

Which of the following statements about heat and work is the proper usage of the terms?

Accepted Answer

Heat and work are processes, not properties that a system possesses. The correct usage involves describing energy transfer between a system and its surroundings. Statement B correctly describes heat transfer, and statement D correctly describes work done by a system on its environment.

Question 19

A lake with 8.0 $\times$ 10⁹kg of water, which has a specific heat of 4180 J/kg · Cº, warms from 10 to 15ºC. The amount of heat transferred to the lake is

Accepted Answer

The answer of A lake with 8.0 $\times$ 10⁹kg...

Question 20

The molar specific heat of copper is 24.5 J/(mol.K). The amount of heat needed to raise 126 g of copper by 2CS1U1P1$\circ$S1S1P0 is

Accepted Answer

To calculate the amount of heat needed, we can use the formula:

q = m × C × ΔT

where q is the amount of heat, m is the mass of the substance, C is the molar specific heat, and ΔT is the change in temperature.

Converting the mass from grams to moles:

126 g Cu / (63.55 g/mol Cu) = 1.98 mol Cu

Substituting the given values into the formula:

q = (1.98 mol Cu) × (24.5 J/(mol·K)) × (2°C)
q = 98.01 J

Rounding to two significant figures, the answer is:

Answer: D) 98 J

Quiz 18: Heat and the First Law of Thermodynamics

Use the following to answer question : -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the solid can be found by

Two liquids, A and B, are mixed together. Liquid A has mass m and was initially at temperature 40^$\circ$C, and liquid B has mass 2m and was initially at temperature 5^$\circ$C. The specific heat of liquid A is 1.5 times that of liquid B. Calculate the final temperature of the mixture.

Body A has twice the mass and three times the specific heat of body B. They are supplied with equal amounts of heat. Body A experiences a temperature change $\Delta$ T. What change in temperature is experienced by body B?

Use the following to answer question : -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the liquid can be found by

Aluminum has a specific heat more than twice that of copper. Identical masses of aluminum and copper, both at 0ºC, are dropped together into a can of hot water. When the system has come to equilibrium,

To raise the temperature of a 2.0-kg piece of metal from 20º to 100ºC, 61.8 kJ of heat is added. What is the specific heat of this metal?

Two liquids, A and B, are mixed together, and the resulting temperature is 22^$\circ$C. If liquid A has mass m and was initially at temperature 35^$\circ$C, and liquid B has mass 3m and was initially at temperature 11^$\circ$C, calculate the ratio of the specific heats of A divided by B.

The specific heat of a substance is a measure of

At liquid ⁴He temperatures (< 4.2 K) , the molar heat capacity heat of most insulators goes as C = AT³, where A is a constant and T is in kelvins. The ratio of the heat required to raise the temperature from 2 K to 3K and from 1 K to 2K is

A system has a heat capacity of 100 J. This means

The quantity of heat absorbed by a body is determined from the formula Q = cm(T_f - T_i) . A certain metal has a specific heat c = 0.21 cal/g Cº and a mass m = 25.6 g. The initial temperature is T_i = 34.6ºC, and the final temperature T_f = 54.6ºC. The quantity of heat absorbed is

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg · Cº) at a temperature of 430ºC is dropped into 48 kg of water. The water is initially at a temperature of 10ºC. With no heat losses to the surroundings, the equilibrium temperature of the iron and water is approximately

For most metals, the heat capacity goes as C = $\gamma$ T + AT³ at low temperatures (~ 100 K) , whereas insulators only has the T³ term. Both $\gamma$ and A are constants. The two terms in the expression for C suggests

A 1.0-kg piece of marble at 100ºC is dropped into 2.5 kg of water at 1.0ºC and the resulting temperature is 7.0ºC. The specific heat of the marble is approximately

Glass beads of mass 100 g and specific heat 0.20 cal/g · Cº are heated to 90ºC and placed in a 300-g glass beaker containing 200 g of water at 20ºC. When equilibrium is reached, the temperature is approximately

Which of the following statements about heat and work is the proper usage of the terms?

A lake with 8.0 $\times$ 10⁹kg of water, which has a specific heat of 4180 J/kg · Cº, warms from 10 to 15ºC. The amount of heat transferred to the lake is

The molar specific heat of copper is 24.5 J/(mol.K) . The amount of heat needed to raise 126 g of copper by 2C^$\circ$ is

Quiz 18: Heat and the First Law of Thermodynamics

Use the following to answer question : -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the solid can be found by

Two liquids, A and B, are mixed together. Liquid A has mass m and was initially at temperature 40 ∘\circ∘ C, and liquid B has mass 2m and was initially at temperature 5 ∘\circ∘ C. The specific heat of liquid A is 1.5 times that of liquid B. Calculate the final temperature of the mixture.

Body A has twice the mass and three times the specific heat of body B. They are supplied with equal amounts of heat. Body A experiences a temperature change Δ\DeltaΔ T. What change in temperature is experienced by body B?

Use the following to answer question : -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The specific heat of the liquid can be found by

Aluminum has a specific heat more than twice that of copper. Identical masses of aluminum and copper, both at 0ºC, are dropped together into a can of hot water. When the system has come to equilibrium,

To raise the temperature of a 2.0-kg piece of metal from 20º to 100ºC, 61.8 kJ of heat is added. What is the specific heat of this metal?

The specific heat of a substance is a measure of

At liquid 4He temperatures (< 4.2 K) , the molar heat capacity heat of most insulators goes as C = AT3, where A is a constant and T is in kelvins. The ratio of the heat required to raise the temperature from 2 K to 3K and from 1 K to 2K is

A system has a heat capacity of 100 J. This means

The quantity of heat absorbed by a body is determined from the formula Q = cm(Tf - Ti) . A certain metal has a specific heat c = 0.21 cal/g Cº and a mass m = 25.6 g. The initial temperature is Ti = 34.6ºC, and the final temperature Tf = 54.6ºC. The quantity of heat absorbed is

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg · Cº) at a temperature of 430ºC is dropped into 48 kg of water. The water is initially at a temperature of 10ºC. With no heat losses to the surroundings, the equilibrium temperature of the iron and water is approximately

For most metals, the heat capacity goes as C = γ\gammaγ T + AT3 at low temperatures (~ 100 K) , whereas insulators only has the T3 term. Both γ\gammaγ and A are constants. The two terms in the expression for C suggests

A 1.0-kg piece of marble at 100ºC is dropped into 2.5 kg of water at 1.0ºC and the resulting temperature is 7.0ºC. The specific heat of the marble is approximately

Glass beads of mass 100 g and specific heat 0.20 cal/g · Cº are heated to 90ºC and placed in a 300-g glass beaker containing 200 g of water at 20ºC. When equilibrium is reached, the temperature is approximately

Which of the following statements about heat and work is the proper usage of the terms?

A lake with 8.0 ×\times× 109 kg of water, which has a specific heat of 4180 J/kg · Cº, warms from 10 to 15ºC. The amount of heat transferred to the lake is

The molar specific heat of copper is 24.5 J/(mol.K) . The amount of heat needed to raise 126 g of copper by 2C ∘\circ∘ is

Two liquids, A and B, are mixed together. Liquid A has mass m and was initially at temperature 40^$\circ$C, and liquid B has mass 2m and was initially at temperature 5^$\circ$C. The specific heat of liquid A is 1.5 times that of liquid B. Calculate the final temperature of the mixture.

Body A has twice the mass and three times the specific heat of body B. They are supplied with equal amounts of heat. Body A experiences a temperature change $\Delta$ T. What change in temperature is experienced by body B?

At liquid ⁴He temperatures (< 4.2 K) , the molar heat capacity heat of most insulators goes as C = AT³, where A is a constant and T is in kelvins. The ratio of the heat required to raise the temperature from 2 K to 3K and from 1 K to 2K is

The quantity of heat absorbed by a body is determined from the formula Q = cm(T_f - T_i) . A certain metal has a specific heat c = 0.21 cal/g Cº and a mass m = 25.6 g. The initial temperature is T_i = 34.6ºC, and the final temperature T_f = 54.6ºC. The quantity of heat absorbed is

For most metals, the heat capacity goes as C = $\gamma$ T + AT³ at low temperatures (~ 100 K) , whereas insulators only has the T³ term. Both $\gamma$ and A are constants. The two terms in the expression for C suggests

A lake with 8.0 $\times$ 10⁹kg of water, which has a specific heat of 4180 J/kg · Cº, warms from 10 to 15ºC. The amount of heat transferred to the lake is

The molar specific heat of copper is 24.5 J/(mol.K) . The amount of heat needed to raise 126 g of copper by 2C^$\circ$ is