Question 1

It is necessary to determine the specific heat of an unknown object.The mass of the object is $\text { 201.0 g. }$ 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 is given by the equation:$$c = \frac{Q}{m\Delta T}$$where:* $c$ is the specific heat (in J/kg∙K)* $Q$ is the heat added (in J)* $m$ is the mass (in kg)* $\Delta T$ is the change in temperature (in K)In this problem, we have:* $Q = 15.0 \mathrm{~J}$* $m = 201.0 \mathrm{~g} = 0.2010 \mathrm{~kg}$* $\Delta T = 10.0 ^\circ \mathrm{C} = 10.0 \mathrm{~K}$Substituting these values into the equation, we get:$$c = \frac{15.0 \mathrm{~J}}{0.2010 \mathrm{~kg} \times 10.0 \mathrm{~K}} = 7.46 \mathrm{~J/kg∙K}$$Therefore, the specific heat of the object is 7.46 J/kg∙K.

Question 2

A cyclic process is carried out on an ideal gas such that it returns to its initial state at the end of a cycle,as shown in the pV diagram in the figure.If the process is carried out in a clockwise sense around the enclosed area,as shown on the figure,then the change of internal energy over the full cycle

Accepted Answer

Since the process completes a full cycle, the internal energy of the gas returns to its initial state. Therefore, the net change in internal energy is zero. The internal energy is a state function, which means that it depends only on the initial and final states of the system and not on the path taken between them. In a cyclic process, the system returns to its initial state, so the change in internal energy is always zero.

Question 3

If 150 kcal of thermal energy 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 = cmΔT

where Q is the thermal energy, c is the specific heat capacity, m is the mass of the material, and ΔT is the change in temperature.

Substituting the given values:

150 kcal = c x 2.0 kg x 400°F

Simplifying:

c = 150 kcal / (2.0 kg x 400°F)

c = 0.1875 kcal / (kg x °C)

Converting kcal to J and °C to K:

c = 0.1875 x 4184 J / (kg x K)

c = 782.4 J / (kg x K)

c = 0.34 kcal / (kg x °C) (rounded to two significant figures)

Therefore, the specific heat capacity of the material is approximately 0.34 kcal/kg ∙ C°. The closest option is (C).

Question 4

A gas is taken through the cycle shown in the pV diagram in the figure.During one cycle,how much work is done by the gas?

Accepted Answer

The answer of A gas is taken through the cycle...

Question 5

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 change the temperature of 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

The first step is to calculate the amount of heat energy required to raise the temperature of 1.0 kg of water from 18°C to 98°C using the specific heat equation:

Q = m * c * ΔT

where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Q = 1.0 kg * 4186 J/(kg ∙ K) * (98°C - 18°C)
Q = 350,280 J

Next, we can use the same equation to calculate the amount of heat energy supplied by the heating coil to the aluminum tea kettle:

Q = m * c * ΔT
Q = 0.5 kg * 900 J/(kg ∙ K) * (98°C - 18°C)
Q = 36,000 J

Now, we can use the equation for power:

P = Q/t

where P is power, Q is energy, and t is time.

We want to solve for t, so we rearrange the equation:

t = Q/P

t = 350,280 J / 500 W + 36,000 J / 500 W
t = 701 s + 72 s
t = 773 s or 12.9 minutes

Therefore, the best choice is C) 12 minutes.

Question 6

A 6.5-g iron meteor hits Earth at a speed of 295 m/s.If its kinetic energy is entirely converted to the thermal energy of 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 given by $ \frac{1}{2}mv^2 $, where $ m $ is the mass in kilograms and $ v $ is the velocity in meters per second. First, convert the mass to kilograms: $ 6.5 \, \text{g} = 0.0065 \, \text{kg} $. Then, calculate the kinetic energy: $ KE = \frac{1}{2} \times 0.0065 \, \text{kg} \times (295 \, \text{m/s})^2 = 281.5125 \, \text{J} $.To find the temperature rise ($ \Delta T $), use the formula $ \Delta T = \frac{Q}{mc} $, where $ Q $ is the thermal energy (equal to the kinetic energy here), $ m $ is the mass in kilograms, and $ c $ is the specific heat capacity in J/kg°C. First, convert the specific heat capacity from cal/kg°C to J/kg°C: $ 113 \, \text{cal/kg°C} = 113 \times 4.186 \, \text{J/kg°C} = 473.018 \, \text{J/kg°C} $.Then, calculate the temperature rise: $ \Delta T = \frac{281.5125 \, \text{J}}{0.0065 \, \text{kg} \times 473.018 \, \text{J/kg°C}} = 92.0 \, \text{C°} $.

Question 7

At room temperature,a typical person transfers energy to the surroundings at the rate of 62 W.If this energy transfer has to be made up by an equivalent food intake,how many kilocalories (food calories)does this person need to consume every day just to make up this energy? (1 cal = 4.186 J)

Accepted Answer

To find the daily energy requirement in kilocalories, first convert the power (62 W) to energy per day: 62 W = 62 J/s. Multiply by the number of seconds in a day (86400 s) to get the total energy in joules: 62 J/s * 86400 s = 5356800 J. Convert joules to kilocalories using the conversion factor (1 kcal = 4186 J): 5356800 J / 4186 J/kcal ≈ 1280 kcal. The closest answer is 1300 kcal.

Question 8

The water flowing over Niagara Falls drops a distance of 50 m.If all the gravitational potential energy of the system water-Earth is converted to thermal energy,by what temperature does the water rise? The specific heat of water is 4186 J/kg ∙ K.

Accepted Answer

The temperature rise can be calculated using the formula for gravitational potential energy (GPE) converted to thermal energy: $\Delta T = \frac{GPE}{m \cdot c}$, where $GPE = m \cdot g \cdot h$, $m$ is the mass of the water (which will cancel out), $g$ is the acceleration due to gravity (9.8 m/s$^2$), $h$ is the height (50 m), and $c$ is the specific heat capacity of water (4186 J/kg·K). Plugging in the values: $\Delta T = \frac{9.8 \cdot 50}{4186} = 0.117$ C°, which rounds to 0.12 C°.

Question 9

A 920-g empty iron kettle is put on a stove.How much thermal energy in joules must be transferred to it 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 thermal energy required to raise the temperature of the kettle is given by:Q = mcΔTwhere 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)Q = 33,900 J

Question 10

A person consumes a snack containing 14 food calories (14 kcal).What is the power this food produces if it is to be "burned off" due to exercise in 6 hours? (1 cal = 4.186 J)

Accepted Answer

First, convert food calories to joules: 14 kcal * 1000 cal/kcal * 4.186 J/cal = 58504 J. Then, divide by time in seconds (6 hours * 3600 seconds/hour) to get power in watts: 58504 J / 21600 s = 2.7 W.

Question 11

On his honeymoon,James Joule attempted to explore the relationships between various forms of energy by measuring the rise of temperature of water which had fallen down a waterfall on Mount Blanc.What maximum temperature rise would one expect for a waterfall with a vertical drop of 20 m? The specific heat of water is 4186 J/kg ∙ K.

Accepted Answer

The answer of On his honeymoon,James Joule attempted to explore...

Question 12

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 converted into internal energyin 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 correct answer is A.The kinetic energy of the hammer is 1/2 * 1.00 kg * (8.50 m/s)^2 = 36.1 J. Half of this energy is converted into internal energy in the nail, so the increase in temperature of the nail is:ΔT = (36.1 J / 2) / (15.0 g * 448 J/kg ∙ K) = 8.1 K

Question 13

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 thermal energy must be added to the gear to raise its temperature from 18°C to 53°C?

Accepted Answer

The thermal energy ($Q$) required to raise the temperature of an object can be calculated using the formula $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For a system consisting of multiple materials, the total energy required is the sum of the energy required for each material. For iron: $Q_{\text{iron}} = 0.10 \, \text{kg} \times 470 \, \text{J/kg} \cdot \text{K} \times (53 - 18) \, \text{K} = 0.10 \times 470 \times 35 = 1645 \, \text{J}$For copper: $Q_{\text{copper}} = 0.16 \, \text{kg} \times 390 \, \text{J/kg} \cdot \text{K} \times (53 - 18) \, \text{K} = 0.16 \times 390 \times 35 = 2184 \, \text{J}$Total energy required: $Q_{\text{total}} = Q_{\text{iron}} + Q_{\text{copper}} = 1645 \, \text{J} + 2184 \, \text{J} = 3829 \, \text{J}$Rounding to the nearest hundred gives approximately 3800 J.

Question 14

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

Accepted Answer

The heat added (Q) is 160,000 J. Using the formula Q = mcΔT, where m is mass (4 kg), c is specific heat capacity (910 J/kg∙K), and ΔT is the change in temperature, we find ΔT = Q/(mc) = 160,000/(4*910) = 44°C. The final temperature is the initial temperature plus ΔT, which is 10°C + 44°C = 54°C.

Question 15

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 16

A certain gas is compressed adiabatically.The amount of work done on the gas is 800 J.What is the change in the internal (thermal)energy of the gas?

Accepted Answer

In an adiabatic process, there is no heat exchange with the surroundings. Therefore, the work done on the gas directly increases its internal energy by the same amount, which is 800 J.

Question 17

A person running in place on an exercise machine for 10 min uses up 17 kcal (food calories).Another person exercises by repeatedly lifting two 2.5-kg weights a distance of 50 cm.How many repetitions of this exercise are equivalent to 10 minutes of running in place? Assume that the person uses negligible energy in letting down the weights after each lift.(1 cal = 4.186 J)

Accepted Answer

The answer of A person running in place on an...

Question 18

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 of the container is negligible,and the specific heat of water is 4.186 × 10³ J/kg ∙ C.

Accepted Answer

To calculate the time required to heat the water, use the formula $Q = mc\Delta T$, where $Q$ is the heat energy in joules, $m$ is the mass in kilograms, $c$ is the specific heat capacity (4.186 J/g°C for water), and $\Delta T$ is the change in temperature. Here, $m = 114.0 \, \text{g} = 0.114 \, \text{kg}$, $c = 4.186 \, \text{J/g°C}$, and $\Delta T = 25°C - 15°C = 10°C$. Thus, $Q = 0.114 \, \text{kg} \times 4.186 \, \text{J/g°C} \times 10°C = 477.084 \, \text{J}$. Since power ($P$) is the rate of energy transfer, and $P = 67.0 \, \text{W} = 67.0 \, \text{J/s}$, the time ($t$) can be found by $t = \frac{Q}{P} = \frac{477.084 \, \text{J}}{67.0 \, \text{J/s}} \approx 7.1 \, \text{s}$.

Question 19

A glass beaker of unknown mass contains $74.0 \mathrm { ml }$ of water.The system absorbs $2000.0 \mathrm { cal }$ of thermal energy and the temperature rises $20.0 ^ { \circ } \mathrm { C }$ as a result.What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C,and that of water is 1.0 cal/g ∙ C°.

Accepted Answer

The total thermal energy absorbed by the system is given as $2000.0 \mathrm { cal }$. This energy is used to raise the temperature of both the water and the glass beaker. The amount of energy required to raise the temperature of the water can be calculated using the formula $q = mc\Delta T$, where $m$ is the mass of the water, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For water, $m = 74.0 \mathrm { g }$ (since $1 \mathrm { ml } = 1 \mathrm { g }$), $c = 1.0 \mathrm { cal/g \cdot ^\circ C }$, and $\Delta T = 20.0 ^\circ \mathrm { C }$. Thus, the energy absorbed by the water is $q_{\text{water}} = 74.0 \mathrm { g } \times 1.0 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C } = 1480.0 \mathrm { cal }$.The remaining energy is absorbed by the glass beaker, which is $2000.0 \mathrm { cal } - 1480.0 \mathrm { cal } = 520.0 \mathrm { cal }$. Using the same formula for the glass beaker, where $c = 0.18 \mathrm { cal/g \cdot ^\circ C }$ and $\Delta T = 20.0 ^\circ \mathrm { C }$, we can solve for the mass of the beaker, $m_{\text{beaker}}$, as follows: $520.0 \mathrm { cal } = m_{\text{beaker}} \times 0.18 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C }$. Solving for $m_{\text{beaker}}$ gives $m_{\text{beaker}} = \frac{520.0 \mathrm { cal }}{0.18 \mathrm { cal/g \cdot ^\circ C } \times 20.0 ^\circ \mathrm { C }} = 144.44 \mathrm { g }$, which rounds to $140 \mathrm { g }$ when considering the options provided.

Question 20

How much thermal energy 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 thermal energy required can be calculated using the formula $q = mc\Delta T$, where $q$ is the thermal energy, $m$ is the mass, $c$ 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, $q = 0.225 \, \text{kg} \times 128 \, \text{J/kg} \cdot \text{K} \times 10.0 \, \text{K} = 288 \, \text{J}$.

Quiz 13: First Law of Thermodynamics