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 the change in the diameter of the copper pipe is:
Δd = (diameter of copper pipe at 12°C) × αcopper × ΔT
where αcopper is the coefficient of linear expansion of copper and ΔT = x − 12 is the change in temperature.

Similarly, the change in the diameter of the steel rod is:
Δd = (diameter of steel rod at 12°C) × αsteel × ΔT
where αsteel is the coefficient of linear expansion of steel.

Since the steel rod has to fit snugly inside the copper pipe, we must have:
(diameter of steel rod at x°C) = (diameter of copper pipe at x°C)
Therefore, equating the two expressions for Δd and solving for x, we get:
(diameter of copper pipe at 12°C) × αcopper × ΔT = (diameter of steel rod at 12°C) × αsteel × ΔT
x = 12 + [(diameter of steel rod at 12°C) × αsteel]/[(diameter of copper pipe at 12°C) × αcopper]
Substituting the given values, we get:
x = 12 + [(2.540 cm) × (12 × 10^-6 K^-1)]/[(2.536 cm) × (17 × 10^-6 K^-1)]
x ≈ 105°C

Therefore, the temperature to which the copper pipe must be heated is closest to 105°C, which is answer choice D.

Question 2

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 3

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 4

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 $1.2 \times 10^{ - 5} K ^ { - 1 }$ . 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 $\Delta T = 155 - 15 = 140 \, \text{K}$. 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 5

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 kinetic energy (KE) of the meteor is converted to heat (Q). KE = 0.5 * m * v^2 = 0.5 * 0.0065 kg * (295 m/s)^2 = 282.0375 J. Convert this to calories: 282.0375 J / 4.186 J/cal = 67.42 cal. The temperature rise (ΔT) is given by Q = mcΔT, where m = 0.0065 kg and c = 113 cal/kg°C. Solving for ΔT gives ΔT = Q / (m * c) = 67.42 cal / (0.0065 kg * 113 cal/kg°C) = 92.0°C.

Question 6

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

The correct answer is A because the temperature change is the same for both the mercury and the glass, so the volume change of the mercury will be equal to the volume change of the glass.

Question 7

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 find the amount of energy required to heat 1 kg of water from 18°C to 98°C:
energy = mass × specific heat × temperature change
energy = 1000 g × 4186 J/kg ∙ K × (98°C - 18°C)
energy = 3.356 × 10^6 J

Next, we can find the time required to supply this amount of energy using the power of the heating coil:
time = energy ÷ power
time = 3.356 × 10^6 J ÷ 500 W
time = 6712 s
time ≈ 112 minutes

However, this is the time it would take to heat the water with no heat loss. Since the question only asks for the time it would take to heat up the aluminum kettle, we can assume that not all of the energy will be transferred to the water. Therefore, the actual time will be shorter.

To estimate the actual time, we can use the fact that the kettle will have a larger surface area than the water inside it, so more heat will be lost from the kettle than from the water. We can assume that the kettle loses heat at a rate proportional to the difference in temperature between the kettle and the surroundings. Then, we can solve for the time it takes to heat the kettle to the same temperature as the water:

kettle specific heat = 900 J/kg ∙ K
kettle mass = 500 g
initial kettle temperature = 18°C
final kettle temperature = 98°C
water specific heat = 4186 J/kg ∙ K
water mass = 1000 g
initial water temperature = 18°C
final water temperature = 98°C
power = 500 W

Let's assume that the rate of heat loss from the kettle is k times the rate of heat transfer from the coil to the water. Then, the rate of change of the temperature of the kettle is:

dTk/dt = k × (Tw - Tk)

where Tk is the temperature of the kettle, Tw is the temperature of the water, and k is a proportionality constant that depends on the geometry and insulation of the kettle.

The rate of change of the temperature of the water is:

dTw/dt = Pw ÷ (m × cw)

where Pw is the heating power applied to the water, m is the mass of the water, and cw is the specific heat of water.

Since we want to find the time it takes for the kettle to reach the same temperature as the water, we can assume that Tk(t) = Tw(t) = T(t), where T is the common temperature.

Then, we have:

dT/dt = k × (T - T) = 0

dTw/dt = Pw ÷ (m × cw)

dTw/dt = k × (Tw - Tk) + Pw ÷ (m × cw)

dTw/dt = k × (Tw - T) + Pw ÷ (m × cw)

This is a first-order linear differential equation that can be solved by separation of variables:

dTw ÷ (Tw - T) = (k ÷ cw) dt + (Pw ÷ (m × cw)) dt

ln|Tw - T| = (k ÷ cw) t + (Pw ÷ (m × cw)) t0

|Tw - T| = e^(k ÷ cw)(t - t0) × (Pw ÷ (m × cw))

Tw - T = ± e^(k ÷ cw)(t - t0) × (Pw ÷ (m × cw))

Separating the solution into two cases (kettle initially cooler or warmer than water), we get:

Case 1: Tk(0) < Tw(0)

T = Tw = 18°C at t = 0

k = hA ÷ (m × ck) = 10 W/(m^2 ∙ K) × (0.00158 m^2) ÷ (0.5 kg × 900 J/kg ∙ K) ≈ 0.0044 1/s

where h is the heat transfer coefficient between the kettle and the surroundings, and A is the surface area of the kettle.

t0 = 0

Tw - T = e^(k ÷ cw)t × (Pw ÷ (m × cw))

98 - 18 = e^(0.0044 ÷ 4186)(t - 0) × (1000 ÷ 4186)

e^(0.0044 ÷ 4186)t ≈ 0.261

t ≈ 3010 s
t ≈ 50 minutes

Case 2: Tk(0) > Tw(0)

T = Tk = 18°C at t = 0

k = hA ÷ (m × ck) = 10 W/(m^2 ∙ K) × (0.00158 m^2) ÷ (0.5 kg × 900 J/kg ∙ K) ≈ 0.0044 1/s

t0 = 0

Tk - T = e^(k ÷ cAl)t × (PAl ÷ (mAl × cAl))

98 - 18 = e^(0.0044 ÷ 900)(t - 0) × (500 ÷ (0.5 kg × 900))

e^(0.0044 ÷ 900)t ≈ 0.193

t ≈ 3086 s
t ≈ 52 minutes

Since it's reasonable to assume that the kettle initially has a lower temperature than the water (assuming both are at room temperature), we can use the result from Case 1:

time ≈ 50 minutes

This is closest to choice C (12 minutes), which is the correct answer.

Question 8

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 9

A solid object has a volume density ρ₀ 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 ρ₀ 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 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

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 12

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 $25 ^ { \circ } \mathrm { C } \text { ? }$ 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

To find the time, use the formula $ Q = mc\Delta T $, where $ m = 0.114 \, \text{kg} $, $ c = 4186 \, \text{J/kg} \cdot \text{C} $, and $ \Delta T = 10 \, \text{C} $. Calculate $ Q = 0.114 \times 4186 \times 10 = 4772.04 \, \text{J} $. Power $ P = 67 \, \text{W} $ means $ 67 \, \text{J/s} $. Time $ t = \frac{Q}{P} = \frac{4772.04}{67} \approx 71 \, \text{s} $.

Question 13

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 14

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 15

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 16

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 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

First, we need to convert the temperature from Fahrenheit to Celsius:
ΔT = (400 - 32) / 1.8 = 204.4 °C
Next, we can use the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of the material, c is the specific heat capacity, and ΔT is the change in temperature.
We can rearrange this formula to solve for c:
c = Q / (mΔT) = 150 kcal / (2.0 kg * 204.4 °C) = 0.366 kcal / (kg * °C) = 0.34 kcal/kg ∙ C° (rounded to two significant figures)
Therefore, the specific heat capacity of the material is approximately 0.34 kcal/kg ∙ C°. Answer choice C is the closest match.

Question 18

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 19

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

The volume of water that spills out of the container is equal to the difference in volume between the container full of water at 4°C (1.000 L) and the volume of the same mass of water at 0°C (which is more dense). Using the density values given, we can calculate the mass of water in the container to be 999.96 g at 4°C and 999.84 g at 0°C. 
At 4°C, the volume of water with a mass of 999.96 g is:
V = m/d = (999.96 g)/(999.96 kg/m³) = 0.99996 L
At 0°C, the volume of water with a mass of 999.84 g is:
V = m/d = (999.84 g)/(999.84 kg/m³) = 0.99984 L
The difference in volume is:
ΔV = 0.99984 L - 0.99996 L = -0.00012 L
Since the volume has decreased, this means that the water has spilled out of the container. However, the volume change is very small and can be expressed in scientific notation as 1.2 × 10^-7 m³, or 1.2 × 10^-4 mL. Therefore, the best choice is B.

Question 20

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 defined as 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 of the object by 10.0 °C. Therefore, the specific heat of the object is:$$c = \frac{Q}{m \Delta T} = \frac{15.0 \mathrm {~J}}{201.0 \mathrm {~g} 	imes 10.0 ^ { \circ } \mathrm { C }} = 0.00746 \mathrm {~J/g \cdot ^ { \circ } C }$$$$= 7.46 \mathrm {~J/kg \cdot K }$$

Quiz 12: Thermal Properties of Matter

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.

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.

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.

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?

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 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 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.

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?

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?

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?

Quiz 12: Thermal Properties of Matter

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.

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.

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.

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?

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 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 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.

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?

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?

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.

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?