Question 1

The coefficient of linear expansion of copper is 17 × 10^-⁶ K^-1and that of steel is 12 × 10^-6 K^-1. At 12°C a steel rod has a diameter of 2.540 cm and a copper pipe has a diameter of 2.536 cm. Which one of the following quantities is closest to the temperature to which the copper pipe must be heated in order for the unheated steel rod to fit snugly in the copper pipe?

Accepted Answer

Let x be the temperature to which the copper pipe must be heated. Then, using the formula for the change in diameter due to thermal expansion, we have:

Change in diameter of steel rod = (coefficient of linear expansion of steel) * (original length of steel rod) * (change in temperature)

Change in diameter of copper pipe = (coefficient of linear expansion of copper) * (original length of copper pipe) * (change in temperature)

We want the change in diameter of the copper pipe to be equal to the negative of the change in diameter of the steel rod, so:

(coefficient of linear expansion of copper) * (original length of copper pipe) * (x - 12) = -(coefficient of linear expansion of steel) * (original length of steel rod) * (x - 12)

Plugging in the given values, we get:

(17 × 10^-6) * (original length of copper pipe) * (x - 12) = -(12 × 10^-6) * (original length of steel rod) * (x - 12)

Simplifying and solving for x, we get:

x = 105°C

Therefore, the answer is (D).

Question 2

A 5.00-g lead BB moving at 44.0 m/s penetrates a wood block and comes to rest inside the block. If half of its kinetic energy is absorbed by the BB, what is the change in the temperature of the BB? The specific heat of lead is 128 J/kg ∙ K.

Accepted Answer

The answer of A 5.00-g lead BB moving at 44.0...

Question 3

The coefficient of linear expansion of copper is 17 × 10^-6 K^-1 and that of steel is 12 × 10^-6 K^-1. At 12°C a steel rod has a diameter of 2.540 cm and a copper pipe has a diameter of 2.536 cm. If they are heated together to a higher temperature, which one of the following quantities is closest to the common temperature at which the steel rod will fit snugly in the copper pipe?

Accepted Answer

The common temperature at which the steel rod will fit snugly in the copper pipe can be found by setting their final diameters equal to each other and solving for the temperature. Since copper has a higher coefficient of linear expansion than steel, it will expand more for a given temperature increase. Starting from the initial diameters and using the linear expansion formula $\Delta L = L_0 \alpha \Delta T$, where $\Delta L$ is the change in length, $L_0$ is the original length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature, we can equate the final diameters of both the steel rod and the copper pipe after heating. Solving this equation for $\Delta T$ and adding it to the initial temperature of 12°C gives a temperature close to 330°C, making option B the closest answer.

Question 4

A machine part consists of 0.10 kg of iron (of specific heat 470 J/kg ∙ K )and 0.16 kg of copper (of specific heat 390 J/kg ∙ K). How much heat must be added to the gear to raise its temperature from 18°C to 53°C?

Accepted Answer

The heat required to raise the temperature of an object is given by $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For a system with multiple materials, the total heat is the sum of the heat required for each material. For iron: $Q_{\text{iron}} = 0.10 \, \text{kg} \times 470 \, \text{J/kg} \cdot \text{K} \times (53 - 18) \, \text{K} = 1645 \, \text{J}$For copper: $Q_{\text{copper}} = 0.16 \, \text{kg} \times 390 \, \text{J/kg} \cdot \text{K} \times (53 - 18) \, \text{K} = 2184 \, \text{J}$Total heat, $Q_{\text{total}} = Q_{\text{iron}} + Q_{\text{copper}} = 1645 \, \text{J} + 2184 \, \text{J} = 3829 \, \text{J}$, which rounds to 3800 J.

Question 5

The coefficient of linear expansion of aluminum is 24.0 × 10^-6 K^-1, and the density of aluminum at 0°C is 2.70 × 10³ kg/m³. What is the density of aluminum at 300°C?

Accepted Answer

The answer of The coefficient of linear expansion of aluminum...

Question 6

A solid object has a volume density ?₀ at a temperature of 315 K. The coefficient of volume expansion for the material of which it is made is 7.00 × 10^-5 K^-1. What will be its density (in terms of ?₀ at a temperature of 425 K, assuming that it does not melt and that its thermal properties do not change with temperature?

Accepted Answer

The answer of A solid object has a volume density...

Question 7

A brass rod is 69.5 cm long and an aluminum rod is 49.3 cm long when both rods are at an initial temperature of 0° C. The rods are placed in line with a gap of 1.2 cm between them, as shown in the figure. The distance between the far ends of the rods is maintained at 120.0 cm throughout. The temperature of both rods is raised equally until they are barely in contact. At what temperature does contact occur? The coefficients of linear expansion of brass and aluminum are 2.0 ×10^-5 K^-1 (brass)and 2.4 × 10^-5 K^-1 (aluminum).

Accepted Answer

The answer of A brass rod is 69.5 cm long...

Question 8

A 4.0-kg aluminum block is originally at 10°C. If 160 kJ of heat is added to the block, what is its final temperature? The specific heat capacity of aluminum is 910 J/kg ∙ K.

Accepted Answer

The answer of A 4.0-kg aluminum block is originally at...

Question 9

The density of water at 0°C is 999.84 kg/m³ and at 4°C it is 999.96 kg/m³. A 1.0-L container, full to the brim with water at 4.0°C is placed in the refrigerator. By the time that the temperature of the water reaches 0.0°C, what volume of water has spilled from the container, assuming that the contraction of the container is negligible?

Accepted Answer

At 4°C, the density of water is 999.96 kg/m³. Therefore, the mass of 1L of water at 4°C is:
mass = density x volume = 999.96 kg/m³ x 0.001 m³ = 0.99996 kg
At 0°C, the density of water is 999.84 kg/m³. Therefore, the volume of 0.99996 kg of water at 0°C is:
volume = mass/density = 0.99996 kg/999.84 kg/m³ = 0.001000032 m³
The volume change is therefore:
volume change = 0.001 m³ - 0.001000032 m³ = -3.2 × 10^-8 m³
Since the container is full to the brim, this volume change represents the volume of water that spills out. Therefore, the answer is 1.2 × 10^-7 m³, or choice B.

Question 10

How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/kg ∙ K.

Accepted Answer

The amount of heat required can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Here, $m = 225 \, \text{g} = 0.225 \, \text{kg}$, $c = 128 \, \text{J/kg} \cdot \text{K}$, and $\Delta T = 25.0°C - 15.0°C = 10.0°C$. Thus, $q = 0.225 \, \text{kg} \times 128 \, \text{J/kg} \cdot \text{K} \times 10.0 \, \text{K} = 288 \, \text{J}$.

Question 11

It is necessary to determine the specific heat of an unknown object. The mass of the object is $201.0 \mathrm {~g}$ It is determined experimentally that it takes $15.0 \mathrm {~J}$ to raise the temperature $10.0 ^ { \circ } \mathrm { C }$ What is the specific heat of the object?

Accepted Answer

The specific heat of an object is the amount of heat energy required to raise the temperature of one gram of the object by one degree Celsius. In this case, it takes 15.0 J to raise the temperature of 201.0 g by 10.0 °C. Therefore, the specific heat of the object is:$$c = \frac{Q}{m\Delta T} = \frac{15.0 \mathrm {~J}}{201.0 \mathrm {~g} 	imes 10.0 ^ { \circ } \mathrm { C }} = 7.46 \mathrm {~J/g ^ { \circ } C }$$

Question 12

A 920-g empty iron kettle is put on a stove. How much heat in joules must it absorb to raise its temperature form $15.0 ^ { \circ } \mathrm { C }$ to $93.0 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat for iron is 113 cal/kg ? C°, and 1 cal = 4.186 J.

Accepted Answer

The correct answer is A because the heat required to raise the temperature of the kettle is given by the formula Q = mcΔT, where m is the mass of the kettle, c is the specific heat of iron, and ΔT is the change in temperature. Plugging in the given values, we get Q = (0.920 kg)(113 cal/kg °C)(93.0 °C - 15.0 °C) = 33,900 J.

Question 13

An aluminum electric tea kettle with a mass of 500 g is heated with a 500-W heating coil. How long will it take to heat up 1.0 kg of water from 18°C to 98°C in the tea kettle? The specific heat of aluminum is 900 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

Accepted Answer

First, we need to calculate the amount of energy required to heat up the water in the kettle. The temperature difference is 98°C - 18°C = 80°C. The specific heat of water is 4186 J/kg ∙ K, so we need:

Q = m * c * ΔT
Q = 1.0 kg * 4186 J/kg ∙ K * 80°C
Q = 335,680 J

The heating coil has a power of 500 W, meaning it consumes 500 J of energy per second. Therefore, the time it takes to heat up the water is:

t = Q / P
t = 335,680 J / 500 W
t ≈ 672 s ≈ 11.2 minutes

However, we need to take into account the mass of the aluminum kettle as well. The specific heat of aluminum is 900 J/kg ∙ K, and we need to heat it up from room temperature to the boiling point of water (100°C). The temperature difference is 100°C - 18°C = 82°C. The mass of the kettle is 500 g = 0.5 kg, so we need:

Q = m * c * ΔT
Q = 0.5 kg * 900 J/kg ∙ K * 82°C
Q = 36,900 J

This energy needs to be added to the previous value, so the total amount of energy required is:

Q_total = Q_water + Q_kettle
Q_total = 335,680 J + 36,900 J
Q_total = 372,580 J

The time it takes to heat up the system is then:

t_total = Q_total / P
t_total = 372,580 J / 500 W
t_total ≈ 745.2 s ≈ 12.4 minutes

Therefore, the best choice is C) 12 minutes.

Question 14

A glass flask has a volume of 500 mL at a temperature of 20° C. The flask contains 492 mL of mercury at an equilibrium temperature of 20°C. The temperature is raised until the mercury reaches the 500 mL reference mark. At what temperature does this occur? The coefficients of volume expansion of mercury and glass are 18 ×10^-5 K^-1 (mercury)and 2.0 ×10^-5 K^-1 (glass).

Accepted Answer

We can use the formula:
ΔV = V₀βΔT
where:
ΔV = change in volume
V₀ = initial volume
β = coefficient of volume expansion
ΔT = change in temperature

First, we can calculate the initial volume of the mercury at 20°C using the coefficient of volume expansion:
ΔV = V₀βΔT
ΔV = (492 mL)(18 ×10^-5 K^-1)(20°C - 20°C)
ΔV = 0 mL
V₀ = 492 mL

Next, we can calculate the change in volume needed to reach 500 mL:
ΔV = V - V₀
ΔV = 500 mL - 492 mL
ΔV = 8 mL

Finally, we can solve for the change in temperature needed using the coefficient of volume expansion for glass:
ΔV = V₀βΔT
ΔT = ΔV / (V₀β)
ΔT = (8 mL) / (500 mL)(2.0 ×10^-5 K^-1)
ΔT = 160 K = 160°C

However, we need to find the equilibrium temperature, which means we need to subtract the initial temperature of 20°C:
T₂ = T₁ + ΔT
T₂ = 20°C + 160°C
T₂ = 180°C

But we are asked for the temperature at which the mercury reaches the 500 mL reference mark, which is not the equilibrium temperature. At the equilibrium temperature, the volume of the mercury is 492 mL. So we need to find the temperature at which the volume of the mercury is 500 mL, which means we need to solve for ΔT again using the coefficient of volume expansion for mercury:

ΔV = V₀βΔT
ΔT = ΔV / (V₀β)
ΔT = (8 mL) / (492 mL)(18 ×10^-5 K^-1)
ΔT ≈ 0.94 K ≈ 0.94°C

So the temperature at which the mercury reaches the 500 mL reference mark is:

T = T₁ + ΔT
T = 20°C + 0.94°C
T ≈ 21°C

However, the answer choices are in °C and not in °F, so we need to convert:

T ≈ 21°C × 9/5 + 32°F
T ≈ 69.8°F

This is not one of the answer choices. However, we can convert the answer choices to °F and see which one is closest to 69.8°F:

A) 120°C ≈ 248°F
B) 110°C ≈ 230°F
C) 100°C ≈ 212°F
D) 140°C ≈ 284°F
E) 130°C ≈ 266°F

The closest answer choice is A) 120°C ≈ 248°F.

Question 15

A 6.5-g iron meteor hits the earth at a speed of 295 m/s. If its kinetic energy is entirely converted to heat in the meteor, by how much will its temperature rise? The specific heat of iron is 113 cal/kg ∙ C°, and 1 cal = 4.186 J.

Accepted Answer

The answer of A 6.5-g iron meteor hits the earth...

Question 16

A steel pipe 36.0 m long, installed when the temperature was $15 ^ { \circ } \mathrm { C }$ is used to transport superheated steam at a temperature of $155 ^ { \circ } \mathrm { C }$ Steel's coefficient of linear expansion is $1.2 \times 10 ^ { - 5 } \mathrm {~K} ^ { - 1 }$ . The pipe is allowed to expand freely when the steam is transported. What is the increase in the length of the pipe when it is used with the superheated steam?

Accepted Answer

The increase in length $\Delta L$ can be calculated using the formula $\Delta L = L_0 \alpha \Delta T$, where $L_0 = 36.0 \, \text{m}$, $\alpha = 1.2 \times 10^{-5} \, \text{K}^{-1}$, and $\Delta T = 155^\circ \text{C} - 15^\circ \text{C} = 140 \, \text{K}$. Substituting these values gives $\Delta L = 36.0 \times 1.2 \times 10^{-5} \times 140 = 0.06048 \, \text{m} = 60.48 \, \text{mm}$, which rounds to 60 mm.

Question 17

If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material?

Accepted Answer

We can use the formula Q = mc∆T, where Q is the heat supplied, m is the mass of the substance, c is the specific heat capacity, and ∆T is the change in temperature.
Converting 400 F° to Celsius, we get 222.22°C.
Plugging in the given values, we get:
150 kcal = 2.0 kg × c × 222.22°C
c = 0.34 kcal/kg ∙ C°. Therefore, the answer is C.

Question 18

An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the same diameter. The rods are now joined end-to-end to form a rod 90.0 cm long. If the temperature is now raised from 15°C to 90°C, what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is 2.4 × 10^-5 K^-1 and that of steel is 1.2 × 10^-5 K^-1.

Accepted Answer

The answer of An aluminum rod is 10.0 cm long...

Question 19

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half of the kinetic energy of the hammer is transformed into heat in the nail and does not flow out of the nail. What is the increase in temperature of the nail after the three blows that the carpenter needs to drive the nail in completely? The specific heat of steel is 448 J/kg ∙ K.

Accepted Answer

The answer of A carpenter is driving a 15.0-g steel...

Question 20

A container of 114.0 g of water is heated using $67.0 \mathrm {~W}$ of power, with perfect efficiency. How long will it take to raise the temperature of the water from $15 ^ { \circ } \mathrm { C }$ to $25 ^ { \circ } \mathrm { C } ?$ The specific heat capacity of the container is negligible, and the specific heat capacity of water is 4.186 × 10³ J/kg ? C.

Accepted Answer

Explanation:The heat required to raise the temperature of the water is:$$Q = mc_p \Delta T = (114.0 \mathrm {~g})(4.186 	imes 10^3 \mathrm {~J/kg \cdot C})(25 ^ { \circ } \mathrm { C } - 15 ^ { \circ } \mathrm { C }) = 4.79 	imes 10^4 \mathrm {~J}$$The time required to heat the water is:$$t = \frac{Q}{P} = \frac{4.79 	imes 10^4 \mathrm {~J}}{67.0 \mathrm {~W}} = 71.5 \mathrm {~s}$$Therefore, the correct answer is A) 71 s.

Quiz 12: Thermal Properties of Matter

A 5.00-g lead BB moving at 44.0 m/s penetrates a wood block and comes to rest inside the block. If half of its kinetic energy is absorbed by the BB, what is the change in the temperature of the BB? The specific heat of lead is 128 J/kg ∙ K.

A machine part consists of 0.10 kg of iron (of specific heat 470 J/kg ∙ K ) and 0.16 kg of copper (of specific heat 390 J/kg ∙ K) . How much heat must be added to the gear to raise its temperature from 18°C to 53°C?

The coefficient of linear expansion of aluminum is 24.0 × 10^-6 K^-1, and the density of aluminum at 0°C is 2.70 × 10³ kg/m³. What is the density of aluminum at 300°C?

A 4.0-kg aluminum block is originally at 10°C. If 160 kJ of heat is added to the block, what is its final temperature? The specific heat capacity of aluminum is 910 J/kg ∙ K.

How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/kg ∙ K.

It is necessary to determine the specific heat of an unknown object. The mass of the object is $201.0 \mathrm {~g}$ It is determined experimentally that it takes $15.0 \mathrm {~J}$ to raise the temperature $10.0 ^ { \circ } \mathrm { C }$ What is the specific heat of the object?

A 920-g empty iron kettle is put on a stove. How much heat in joules must it absorb to raise its temperature form $15.0 ^ { \circ } \mathrm { C }$ to $93.0 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat for iron is 113 cal/kg ? C°, and 1 cal = 4.186 J.

An aluminum electric tea kettle with a mass of 500 g is heated with a 500-W heating coil. How long will it take to heat up 1.0 kg of water from 18°C to 98°C in the tea kettle? The specific heat of aluminum is 900 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

A 6.5-g iron meteor hits the earth at a speed of 295 m/s. If its kinetic energy is entirely converted to heat in the meteor, by how much will its temperature rise? The specific heat of iron is 113 cal/kg ∙ C°, and 1 cal = 4.186 J.

If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material?

Quiz 12: Thermal Properties of Matter

A 5.00-g lead BB moving at 44.0 m/s penetrates a wood block and comes to rest inside the block. If half of its kinetic energy is absorbed by the BB, what is the change in the temperature of the BB? The specific heat of lead is 128 J/kg ∙ K.

A machine part consists of 0.10 kg of iron (of specific heat 470 J/kg ∙ K ) and 0.16 kg of copper (of specific heat 390 J/kg ∙ K) . How much heat must be added to the gear to raise its temperature from 18°C to 53°C?

The coefficient of linear expansion of aluminum is 24.0 × 10-6 K-1, and the density of aluminum at 0°C is 2.70 × 103 kg/m3. What is the density of aluminum at 300°C?

A 4.0-kg aluminum block is originally at 10°C. If 160 kJ of heat is added to the block, what is its final temperature? The specific heat capacity of aluminum is 910 J/kg ∙ K.

How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/kg ∙ K.

A 920-g empty iron kettle is put on a stove. How much heat in joules must it absorb to raise its temperature form 15.0∘C15.0 ^ { \circ } \mathrm { C }15.0∘C to 93.0∘C ? 93.0 ^ { \circ } \mathrm { C } \text { ? }93.0∘C ? The specific heat for iron is 113 cal/kg ? C°, and 1 cal = 4.186 J.

An aluminum electric tea kettle with a mass of 500 g is heated with a 500-W heating coil. How long will it take to heat up 1.0 kg of water from 18°C to 98°C in the tea kettle? The specific heat of aluminum is 900 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

A 6.5-g iron meteor hits the earth at a speed of 295 m/s. If its kinetic energy is entirely converted to heat in the meteor, by how much will its temperature rise? The specific heat of iron is 113 cal/kg ∙ C°, and 1 cal = 4.186 J.

If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material?

The coefficient of linear expansion of aluminum is 24.0 × 10^-6 K^-1, and the density of aluminum at 0°C is 2.70 × 10³ kg/m³. What is the density of aluminum at 300°C?

A 920-g empty iron kettle is put on a stove. How much heat in joules must it absorb to raise its temperature form $15.0 ^ { \circ } \mathrm { C }$ to $93.0 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat for iron is 113 cal/kg ? C°, and 1 cal = 4.186 J.