Question 1

How many grams of ice at -13°C must be added to 711 grams of water that is initially at a temperature of 87°C to produce water at a final temperature of 10.0°C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 J/kg·C° and of ice is 2050 J/kg·C°. For water the normal melting point is 0.00°C and the heat of fusion is 334 × 10³ J/kg. The normal boiling point is 100°C and the heat of vaporization is 2.26 × 10⁶ J/kg.

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Question 2

Heat is added to a 2.0 kg piece of ice at a rate of 793.0 kW. How long will it take for the ice to melt if it was initially at 0.00°C? (The latent heat of fusion for water is 334 kJ/kg and its latent heat of vaporization is 2260 kJ/kg.)

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Question 3

An 80-g aluminum calorimeter contains 380 g of water at an equilibrium temperature of 20°C. A 120-g piece of metal, initially at 352°C, is added to the calorimeter. The final temperature at equilibrium is 32°C. Assume there is no external heat exchange. The specific heats of aluminum and water are 910 J/kg·K and 4190 J/kg·K, respectively. The specific heat of the metal is closest to

Accepted Answer

First, we need to use the principle of conservation of energy:
The heat lost by the hot metal = the heat gained by the aluminum calorimeter and the water
This can be represented as:
(metal mass)(metal specific heat)(final temperature - initial temperature) = (aluminum calorimeter mass)(aluminum specific heat)(final temperature - initial temperature) + (water mass)(water specific heat)(final temperature - initial temperature)
Plugging in the given values and solving for the specific heat of the metal:
(120 g)(C)(352°C - 32°C) = (80 g)(910 J/kg·K)(32°C - 20°C) + (380 g)(4190 J/kg·K)(32°C - 20°C)
Where C represents the specific heat of the metal.
Simplifying and solving for C gives:
C ≈ 520 J/kg·K
Therefore, the closest option is A) 520 J/kg·K.

Question 4

A substance has a melting point of 20°C and a heat of fusion of 3.9 x 10⁴ J/kg. The boiling point is 150°C and the heat of vaporization is 7.8 x 10⁴ J/kg at a pressure of 1.0 atm. The specific heats for the solid, liquid, and gaseous phases are 600 J/(kg∙K), 1000 J/(kg∙K), and 400 J/(kg∙K), respectively. The quantity of heat required to raise the temperature of 3.80 kg of the substance from -61°C to 128°C, at a pressure of 1.0 atm, is closest to

Accepted Answer

The correct answer is A because the heat required to raise the temperature of 3.80 kg of the substance from -61°C to 128°C, at a pressure of 1.0 atm, is closest to 620 kJ.

Question 5

If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g∙K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g.

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Question 6

What is the steady state rate of heat flow through a pane of glass that is 40.0 cm by 30.0 cm with a thickness of 4.00 mm when the outside temperature of the glass is -10.0°C and its inside temperature is 25.0°C? The thermal conductivity of glass is 0.105 W/(m∙K), the specific heat of glass is 0.180 cal/(g∙°C), and 1 cal = 4.190 J.

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Question 7

If you add 700 kJ of heat to 700 g of water at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 2.26 × 10⁶ J/kg and its specific heat is is 4190 J/(kg∙K).

Accepted Answer

First, calculate the energy needed to raise the temperature of water from 70°C to 100°C:  Q = mcΔT = 0.7 kg * 4190 J/(kg∙K) * (100°C - 70°C) = 87990 J.  Convert 700 kJ to J: 700 kJ = 700,000 J.  Energy left for vaporization: 700,000 J - 87990 J = 612010 J.  Mass vaporized = Energy left / Latent heat = 612010 J / 2.26 × 10^6 J/kg ≈ 0.271 kg = 271 g.  Initial mass = 700 g, so mass left = 700 g - 271 g = 429 g.

Question 8

Two experimental runs are performed to determine the calorimetric properties of an alcohol that has a melting point of -10°C. In the first run, a 200-g cube of frozen alcohol, at the melting point, is added to 300 g of water at 20°C in a styrofoam container. When thermal equilibrium is reached, the alcohol-water solution is at a temperature of 5.0°C. In the second run, an identical cube of alcohol is added to 500 g of water at 20°C and the temperature at thermal equilibrium is 10°C. The specific heat of water is 4190 J/kg·K. Assume that no heat is exchanged with the styrofoam container and with the surroundings. The heat of fusion of the alcohol is closest to

Accepted Answer

First, we can use the equation:

Q = mCΔT

where Q is the heat exchanged, m is the mass, C is the specific heat, and ΔT is the change in temperature.

For the first run, we can find the heat released by the alcohol when it melts:

Q1 = (0.2 kg)(-33.15 kJ/kg) = -6.63 kJ

where we use the heat of fusion of water (-33.15 kJ/kg) since the alcohol is at its melting point.

This heat is then absorbed by the water as it warms up from 20°C to 5.0°C:

Q2 = (0.3 kg)(4190 J/kg·K)(5.0°C - 20°C) = -6.285 kJ

The negative sign means that heat is being lost by the water. Setting Q1 = Q2, we can solve for the heat of fusion:

-6.63 kJ = -6.285 kJ + (0.2 kg)Hf

Hf = 6.225 kJ/kg

For the second run, we follow the same steps:

Q1 = (0.2 kg)(-33.15 kJ/kg) = -6.63 kJ

Q2 = (0.5 kg)(4190 J/kg·K)(10°C - 20°C) = -20.95 kJ

Setting Q1 = Q2, we can solve for the heat of fusion:

-6.63 kJ = -20.95 kJ + (0.2 kg)Hf

Hf = 6.32 kJ/kg

Taking the average of these two values, we get:

Hf = 6.275 kJ/kg

which is closest to choice B, 6.3 × 10^3 J/kg.

Question 9

A 406.0 kg copper bar is put into a smelter for melting. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? (The specific heat for copper is 386 J/kg•K, the heat of fusion for copper is 205 kJ/kg, and its melting point is 1357 K.)

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Question 10

A copper cylinder with a mass of 125 g and temperature of 345°C is cooled by dropping it into a glass beaker containing 565 g of water initially at 20.0°C. The mass of the beaker is 50.0 g and the specific heat of the glass is 840 J/kg∙K. What is the final equilibrium temperature of the system, assuming the cooling takes place very quickly, so that no energy is lost to the air? The specific heat of copper is 385 J/kg∙K and that of water is 4190 J/kg∙K.

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Question 11

A person pours 330 g of water at 45°C into an 855-g aluminum container with an initial temperature of 10°C. The specific heat of aluminum is 900 J/(kg∙K) and that of water is 4190 J/(kg∙K). What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

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Question 12

Heat is added to a pure substance in a closed container at a constant rate. The figure shows a graph of the temperature of the substance as a function of time. If L_f = latent heat of fusion and L_v = latent heat of vaporization, what is the value of the ratio L_v / L_f for this substance?

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Question 13

A block of ice at 0.000°C is added to a well-insulated 147-g aluminum calorimeter cup that holds 200 g of water at 10.0°C. The water and aluminum cup are in thermal equilibrium, and the specific heat of aluminum is 910 J/(kg∙K). If all but 2.00 g of ice melt, what was the original mass of the block of ice? The specific heat of water is 4190 J/(kg∙K), its latent heat of fusion is 334 kJ/kg, and its latent heat of vaporization is 2260 kJ/kg.

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Question 14

A 400-g piece of metal at 120.0°C is dropped into a cup containing 450 g of water at 15.0°C. The final temperature of the system is measured to be 40.0°C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings or the cup? The specific heat of water is 4190 J/(kg∙K).

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Question 15

Two experimental runs are performed to determine the calorimetric properties of an alcohol that has a melting point of -10°C. In the first run, a 200-g cube of frozen alcohol, at the melting point, is added to 300 g of water at 20°C in a styrofoam container. When thermal equilibrium is reached, the alcohol-water solution is at a temperature of 5.0°C. In the second run, an identical cube of alcohol is added to 500 g of water at 20°C and the temperature at thermal equilibrium is 10°C. The specific heat of water is 4190 J/kg·K. Assume that no heat is exchanged with the styrofoam container and with the surroundings. The specific heat of the alcohol is closest to

Accepted Answer

In the first run, the heat lost by the alcohol is gained by the water:
$$Q_{lost\ by\ alcohol}=Q_{gained\ by\ water}$$
$$(m_{alcohol}	imes C_{alcohol}	imes \Delta T)=(m_{water}	imes C_{water}	imes \Delta T)$$
$$(0.2\ kg)	imes C_{alcohol}	imes (15\ ^{\circ}C)= (0.3\ kg)	imes (4190\ J/kg	imes K)	imes (15\ ^{\circ}C)$$ 
$$C_{alcohol}=2100\ J/kg	imes K$$

Similarly, in the second run:
$$(m_{alcohol}	imes C_{alcohol}	imes \Delta T)=(m_{water}	imes C_{water}	imes \Delta T)$$
$$(0.2\ kg)	imes C_{alcohol}	imes (10\ ^{\circ}C)= (0.5\ kg)	imes (4190\ J/kg	imes K)	imes (10\ ^{\circ}C)$$ 
$$C_{alcohol}=1900\ J/kg	imes K$$
Taking the average of the two values we get:
$$C_{alcohol}=(1900+2100)/2=2000\ J/kg	imes K$$
Therefore, the closest option is (C) 2100 J/kg • K.

Question 16

A person makes ice tea by adding ice to 1.8 kg of hot tea, initially at 80°C. How many kilograms of ice, initially at 0.00°C, are required to bring the mixture to 10°C? The heat of fusion of ice is 334 kJ/kg, and we can assume that tea has essentially the same thermal properties as water, so its specific heat is 4190 J/(kg∙K).

Accepted Answer

We can use the equation Q = mcdeltaT to solve this problem, where Q is the heat exchanged, m is the mass of the substance, c is the specific heat, and deltaT is the change in temperature.

First, we need to find the initial heat of the hot tea:

Q1 = (1.8 kg)(4190 J/(kg*K))(80°C - 10°C) 
Q1 = 444,312 J

Next, we need to find the amount of heat that the ice will absorb as it melts and warms up to 10°C:

Q2 = (mice)(334 kJ/kg) + (mice)(4190 J/(kg*K))(10°C - 0°C) 
Q2 = (mice)(334000 J/kg) + (mice)(41900 J/(kg*K))(10°C - 0°C) 
Q2 = (mice)(459400 J)

Since the heat lost by the hot tea is equal to the heat gained by the ice, we can set Q1 equal to Q2:

444,312 J = (mice)(459400 J) 
mice = 0.967 kg

Therefore, the person needs to add 0.967 kg of ice, which is closest to choice C) 1.4 kg.

Question 17

A 905-g meteor impacts the earth at a speed of 1629 m/s. If all of its energy is entirely converted to heat in the meteor, what will be the resulting temperature rise of the meteor, assuming it does not melt? The specific heat for the meteor material is 472 J/kg∙K, which is about the same as that of iron.

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Question 18

Under steady state conditions, a piece of wood 350 mm by 350 mm and 15 mm thick conducts heat through its thickness and loses no appreciable heat through its well-insulated sides. The rate of heat flow is measured to be 14.0 W when the temperature difference across its thickness is 28 C°. Determine the thermal conductivity of this wood.

Accepted Answer

We can use the formula for steady-state heat transfer through a plane wall:

Q = kAΔT/d

where Q is the rate of heat flow, k is the thermal conductivity, A is the area of the wall, ΔT is the temperature difference across the wall, and d is the thickness of the wall.

Plugging in the given values, we get:

14.0 = k(0.35 x 0.35)(28)/0.015

Solving for k, we get:

k = 0.061 W/(m∙C°)

Therefore, the answer is D.

Question 19

A 200-g metal container, insulated on the outside, holds 100 g of water in thermal equilibrium at 22.00°C. A 21-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.00°C. Assume there is no heat exchange with the surroundings. For water, the specific heat is 4190 J/kg · K and the heat of fusion is 3.34 × 10⁵ J/kg. The specific heat for the metal is closest to

Accepted Answer

First, we can calculate the amount of heat released by the ice as it melts:
$q_{ice} = m_{ice} \cdot L_f = 21	ext{ g} \cdot 334	ext{ J/g} = 7026	ext{ J}$
where $L_f$ is the heat of fusion.

This heat is absorbed by the water as the ice melts and by the water and container as the resulting water cools from 22.00°C to 15.00°C:
$q_{water+container} = (m_{water} c_{water} + m_{container} c_{container}) \Delta T$
where $\Delta T = 22.00	ext{°C} - 15.00	ext{°C} = 7.00	ext{°C}$.

We know the mass of the water and the ice, but not the mass or specific heat of the container. However, we can use the fact that the container is insulated to assume that there is no heat exchange with the surroundings. This means that the total amount of heat in the system (water, ice, and container) is conserved:

$q_{ice} = -q_{water+container}$

Plugging in the values we know and solving for $c_{container}$, we get:

$c_{container} = \frac{q_{water+container}}{m_{container} \Delta T} = \frac{-q_{ice}}{m_{container} \Delta T} = \frac{-7026	ext{ J}}{200	ext{ g} \cdot 7.00	ext{°C}} \approx 3850	ext{ J/kg} \cdot 	ext{K}$

Therefore, the closest choice for the specific heat of the container is (A) 3850 J/kg ∙ K.

Question 20

A substance has a melting point of 20°C and a heat of fusion of 3.5 × 10⁴ J/kg. The boiling point is 150°C and the heat of vaporization is 7.0 × 10⁴ J/kg at a pressure of 1.0 atm. The specific heats for the solid, liquid, and gaseous phases are 600 J/(kg∙K), 1000 J/(kg∙K), and 400 J/(kg∙K), respectively. The quantity of heat given up by 0.50 kg of the substance when it is cooled from 170°C to 88°C, at a pressure of 1.0 atmosphere, is closest to

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Quiz 17: Work, Heat, and the First Law of Thermodynamics

Heat is added to a 2.0 kg piece of ice at a rate of 793.0 kW. How long will it take for the ice to melt if it was initially at 0.00°C? (The latent heat of fusion for water is 334 kJ/kg and its latent heat of vaporization is 2260 kJ/kg.)

If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g∙K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g.

If you add 700 kJ of heat to 700 g of water at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 2.26 × 10⁶ J/kg and its specific heat is is 4190 J/(kg∙K) .

A person pours 330 g of water at 45°C into an 855-g aluminum container with an initial temperature of 10°C. The specific heat of aluminum is 900 J/(kg∙K) and that of water is 4190 J/(kg∙K) . What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

Heat is added to a pure substance in a closed container at a constant rate. The figure shows a graph of the temperature of the substance as a function of time. If L_f = latent heat of fusion and L_v = latent heat of vaporization, what is the value of the ratio L_v / L_f for this substance?

Quiz 17: Work, Heat, and the First Law of Thermodynamics

Heat is added to a 2.0 kg piece of ice at a rate of 793.0 kW. How long will it take for the ice to melt if it was initially at 0.00°C? (The latent heat of fusion for water is 334 kJ/kg and its latent heat of vaporization is 2260 kJ/kg.)

If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g∙K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g.

If you add 700 kJ of heat to 700 g of water at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 2.26 × 106 J/kg and its specific heat is is 4190 J/(kg∙K) .

A person pours 330 g of water at 45°C into an 855-g aluminum container with an initial temperature of 10°C. The specific heat of aluminum is 900 J/(kg∙K) and that of water is 4190 J/(kg∙K) . What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

Heat is added to a pure substance in a closed container at a constant rate. The figure shows a graph of the temperature of the substance as a function of time. If Lf = latent heat of fusion and Lv = latent heat of vaporization, what is the value of the ratio Lv / Lf for this substance?

If you add 700 kJ of heat to 700 g of water at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 2.26 × 10⁶ J/kg and its specific heat is is 4190 J/(kg∙K) .

Heat is added to a pure substance in a closed container at a constant rate. The figure shows a graph of the temperature of the substance as a function of time. If L_f = latent heat of fusion and L_v = latent heat of vaporization, what is the value of the ratio L_v / L_f for this substance?