Question 1

In typical calorimetry problems,what is the usual conservation law being used?
A)energy
B)heat
C)temperature
D)None since nothing is being conserved.

Accepted Answer

The usual conservation law being used in calorimetry problems is the conservation of energy. This means that the total energy in a system before a reaction or change is equal to the total energy afterwards. In calorimetry, we measure the change in energy in a system by measuring the heat transfer between the system and its surroundings. This heat transfer can be calculated using the specific heat of the materials involved and the change in temperature. Therefore, energy conservation is fundamental in calorimetry problems to ensure that the calculated values are accurate.

Question 2

A swimming pool heater has to be able to raise the temperature of the 50 000 gallons of water in the pool by 12.0 C°.How many kilowatt-hours of energy are required? (One gallon of water has a mass of approximately 3.8 kg and the specific heat of water is 4,186 J/kg⋅°C. )
A)6.98E+2 kWh
B)2.65E+3 kWh
C)2.21E+2 kWh
D)2.65E+4 kWh
E)2.65E+2 kWh

Accepted Answer

The total heat energy required to raise the temperature of the pool by 12.0 C° can be calculated using the formula:

Q = m*Cp*ΔT

where 
m = mass of water = volume of water * density of water 
= (50,000 gallons) * (3.8 kg/gallon) 
= 190,000 kg

Cp = specific heat of water = 4,186 J/kg⋅°C

ΔT = temperature change = 12.0 C°

Therefore, 
Q = (190,000 kg) * (4,186 J/kg⋅°C) * (12.0 C°) 
= 9,969,120,000 J

Converting J to kWh, we get:

9,969,120,000 J * (1 kWh / 3.6E+6 J) 
= 2.77E+3 kWh

Therefore, the answer is B) 2.65E+3 kWh, which is the closest option.

Question 3

A 3.00-kg copper rod is 45.0 cm long at 24.8°C.If 45 000 J are transferred to the rod by heat,what is its change in length? c_copper = 387 J/kg⋅°C and α_copper = 17.0 × 10−⁶/°C. A)34.4 cm B)0.0297 cm C)0.133 cm D)0.0445 cm E)More information is needed.

Accepted Answer

The change in length can be calculated using the formula for thermal expansion: ΔL = αL0ΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ΔT is the change in temperature. First, we need to find ΔT using the heat transferred (Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature). Rearranging for ΔT gives ΔT = Q/(mc). Substituting the given values: ΔT = 45000 J / (3.00 kg * 387 J/kg°C) = 38.76°C. Then, using ΔL = αL0ΔT: ΔL = 17.0 × 10^-6 /°C * 45.0 cm * 38.76°C = 0.0297 cm.

Question 4

Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C.Thermal equilibrium is reached at 30°C.The specific heat of the solid
A)is equal to that of the fluid.
B)is less than that of the fluid.
C)is more than that of the fluid.
D)cannot be compared to that of a material in a different phase.

Accepted Answer

The specific heat of the solid is more than that of the fluid because the solid, having a smaller mass, requires less heat to change its temperature compared to the larger mass of the fluid, indicating a higher specific heat capacity.

Question 5

Which of the following has the highest specific heat?
A)ice
B)water
C)steam
D)They are all water,so they all have the same specific heat.

Accepted Answer

Water has the highest specific heat among the given options. This is because it takes a lot of energy to raise the temperature of water due to hydrogen bonding, which allows it to absorb more heat energy without a significant rise in temperature.  Ice and steam have lower specific heats than liquid water.

Question 6

What is the temperature increase of 3.00 kg of water when heated by a 900-W immersion heater for 9.00 min? (c_w = 4 186 J/kg⋅°C) A)77.4°C B)48.4°C C)38.7°C D)19.4°C E)0.645°C

Accepted Answer

The amount of heat transferred to the water can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of water, c is the specific heat capacity, and ΔT is the temperature increase.

We have:

m = 3.00 kg
c = 4 186 J/kg⋅°C
ΔT = (900 W) × (9.00 min) × (60 s/min) / (3.00 kg × 4 186 J/kg⋅°C) = 38.7°C

Therefore, the temperature increase of the water is 38.7°C, so the answer is choice C.

Question 7

A slice of bread contains about 125 kcal.If specific heat of a person were 1.00 kcal/kg⋅°C,by how many °C would the temperature of a 65.0-kg person increase if all the energy in the bread were converted to heat?
A)3.85°C
B)0.96°C
C)1.92°C
D)2.40°C
E)1.44°C

Accepted Answer

The amount of energy in the slice of bread is 125 kcal, which is the same as 125,000 calories. If all of this energy were converted to heat and absorbed by the person, the heat gained by the person would be:

Heat gained = energy in food = 125,000 calories

Assuming the specific heat of the person is 1.00 kcal/kg⋅°C, we can use the formula for specific heat:

Heat gained = (mass of person) x (specific heat) x (change in temperature)

We can rearrange this formula to solve for the change in temperature:

Change in temperature = Heat gained / [(mass of person) x (specific heat)]

Substituting the values given:

Change in temperature = 125,000 cal / [(65.0 kg) x (1.00 kcal/kg⋅°C)]

Change in temperature = 1923.08 °C

Therefore, the temperature of the person would increase by 1.92°C if all the energy in the bread were converted to heat. The best choice is (C).

Question 8

A piece of copper of mass 110 g is being drilled through with a ¹/₂" electric drill.The drill operates at 25.0 W and takes 25.0 s to bore through the copper.If all the energy from the drill heats the copper,find the copper's increase in temperature.c_copper = 387 J/kg⋅°C. A)18.4 C° B)23.5 C° C)14.7 C° D)22.0 C° E)29.4 C°

Accepted Answer

The energy transferred to the copper is equal to the power of the drill times the time it operates, which is $25.0\, \text{W} \times 25.0\, \text{s} = 625\, \text{J}$. The mass of the copper is $0.110\, \text{kg}$, and the specific heat capacity of copper is $387\, \text{J/kg}\cdot\text{°C}$. The increase in temperature ($\Delta T$) can be found using the formula $\Delta T = \frac{Q}{m \cdot c}$, where $Q$ is the heat energy transferred, $m$ is the mass, and $c$ is the specific heat capacity. Substituting the given values, $\Delta T = \frac{625}{0.110 \cdot 387} \approx 14.7\, \text{°C}$.

Question 9

A solar heating system has a 25.0% conversion efficiency;the solar radiation incident on the panels is 800 W/m².What is the increase in temperature of 25.0 kg of water in a 1.00-h period by a 3.40-m²-area collector? (c_w = 4 186 J/kg⋅°C) A)93.6°C B)6.88°C C)23.4°C D)46.8°C E)17.5°C

Accepted Answer

The rate of energy collection is:$$P=A\eta I=(3.40	ext{ m}^2)(0.250)(800	ext{ W/m}^2)=680	ext{ W}$$The increase in temperature is:$$\Delta T=\frac{Pt}{mc_w}=\frac{(680	ext{ W})(3600	ext{ s})}{(25.0	ext{ kg})(4186	ext{ J/kg}\cdot\degree	ext{C})}=\boxed{23.4\degree	ext{C}}$$

Question 10

A 0.200-kg aluminum plate,initially at 20°C,slides down a 13.0-m-long surface,inclined at a 25.0° angle to the horizontal.The force of kinetic friction exactly balances the component of gravity down the plane so that the plate,once started,glides down at constant velocity.If 90% of the mechanical energy of the system is absorbed by the aluminum,what is its temperature increase at the bottom of the incline? (Specific heat for aluminum is 900 J/kg⋅°C. )
A)0.115 C°
B)0.0538 C°
C)0.269 C°
D)0.78 C°
E)48.5 C°

Accepted Answer

First, we need to find the initial mechanical energy of the system. Using conservation of energy, we have:

Initial mechanical energy = mgh = (0.200 kg)(9.81 m/s^2)(sin 25°)(13.0 m) = 24.1 J

Since 90% of this energy is absorbed by the aluminum plate, the plate gains an amount of thermal energy equal to 0.9 times the initial mechanical energy:

Thermal energy gained by plate = 0.9(24.1 J) = 21.7 J

To find the temperature increase, we divide the thermal energy gained by the specific heat of aluminum and the mass of the plate:

ΔT = (21.7 J)/(0.200 kg)(900 J/kg⋅°C) = 0.0538 °C

Therefore, the temperature of the aluminum plate increases by approximately 0.0538°C at the bottom of the incline. The correct answer is B.

Question 11

A 4.00-g lead bullet is traveling at a speed of 200 m/s when it embeds in a wood post.If we assume that half of the resultant heat energy generated remains with the bullet,what is the increase in temperature of the embedded bullet? (specific heat of lead = 0.0305 kcal/kg⋅°C,1 kcal = 4 186 J)
A)78.3 °C
B)157 °C
C)313 °C
D)39.2 °C
E)97.9 °C

Accepted Answer

The kinetic energy of the bullet is converted into heat energy. The kinetic energy is (1/2)mv^2 = (1/2)(0.004 kg)(200 m/s)^2 = 80 J. Half of this energy remains with the bullet, so 40 J is used to increase its temperature. Convert 40 J to kcal: 40 J / 4186 J/kcal = 0.00956 kcal. The temperature increase is given by Q = mcΔT, where m = 0.004 kg and c = 0.0305 kcal/kg°C. Solving for ΔT gives ΔT = Q/(mc) = 0.00956 kcal / (0.004 kg * 0.0305 kcal/kg°C) = 78.3 °C.

Question 12

A machine gear consists of 0.14 kg of iron and 0.19 kg of copper.How much total heat is generated in the part if its temperature increases by 35.0 C°? (Specific heats of iron and copper are 450 and 390 J/kg⋅°C,respectively. )
A)4.5E+3 J
B)4.8E+3 J
C)5.2E+3 J
D)2.2E+3 J
E)2.6E+3 J

Accepted Answer

First, we need to calculate the heat generated separately for iron and copper, and then add them to get the total heat generated.

Heat generated by iron = mass x specific heat x temperature change 
= 0.14 kg x 450 J/kg⋅°C x 35.0 C° 
= 2205 J

Heat generated by copper = mass x specific heat x temperature change 
= 0.19 kg x 390 J/kg⋅°C x 35.0 C° 
= 2565 J

Total heat generated = Heat generated by iron + Heat generated by copper 
= 2205 J + 2565 J 
= 4770 J

Rounding off to 2 significant figures, we get 4.8E+3 J as the answer. Therefore, the best choice is B.

Question 13

In the text there is a table of coefficients of linear (α)and volume (β)expansions due to temperature changes.For the solid materials in this list,there is a relationship between α and β which allows you to find the value of β if you know α.Which of the following best describes that relationship? A)α = β³ B)3α = β C)α = 3β D)α³ = β

Accepted Answer

The relationship between the linear expansion coefficient (α) and the volume expansion coefficient (β) for solid materials is given by β = 3α. This is because volume expansion is roughly three times the linear expansion in each dimension (length, width, height) for isotropic materials.

Question 14

A waterfall is 128 m high.What is the increase in water temperature at the bottom of the falls if all the initial potential energy goes into heating the water? (g = 9.8 m/s²,c_w = 4 186 J/kg⋅°C) A)0.15°C B)0.30°C C)0.90°C D)0.37°C E)0.60°C

Accepted Answer

The potential energy converted to heat is mgh = mcΔT. Solving for ΔT gives ΔT = gh/c. Substituting the given values, ΔT = (9.8 m/s² * 128 m) / 4186 J/kg°C = 0.30°C.

Question 15

A solar heated house loses about 5.5 × 10⁷ cal through its outer surfaces on a typical 24-h winter day.What mass of storage rock is needed to provide this amount of heat if it is brought up to initial temperature of 58°C by the solar collectors and the house is maintained at 23°C? (Specific heat of rock is 0.21 cal/g⋅°C. ) A)1.4E+3 kg B)7.5E+4 kg C)7.5E+3 kg D)7.5E+2 kg E)1.6E+3 kg

Accepted Answer

We can use the formula:

Q = mcΔT

where Q is the amount of heat needed, m is the mass of the storage rock, c is the specific heat, and ΔT is the temperature change.

We can first convert the heat loss to Joules:

5.5 × 10^17 cal = 5.5 × 10^17 × 4.184 J/cal = 2.3 × 10^18 J

Next, we can calculate the temperature difference:

ΔT = (58°C - 23°C) = 35°C

Now we can solve for the mass:

m = Q / (cΔT) = (2.3 × 10^18 J) / (0.21 cal/g°C × 4.184 J/cal × 35°C) ≈ 7.5 × 10^3 kg

Therefore, the best answer is (C) 7.5E+3 kg.

Question 16

A hot (85.0°C)lump of metal has a mass of 210 g and a specific heat of 0.250 cal/g⋅°C.John drops the metal into a 345-g calorimeter containing 95 g of water at 21.3°C.The calorimeter is constructed of a material that has a specific heat of 0.100 cal/ g⋅°C.When equilibrium is reached,what will be the final temperature? c_water = 1.00 cal/g⋅°C. A)35.6°C B)63.9°C C)49.6°C D)39.7°C E)29.4°C

Accepted Answer

To find the final temperature, we use the principle of conservation of energy, where the heat lost by the metal and calorimeter equals the heat gained by the water. The equation for heat is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is the change in temperature.Let $T_f$ be the final temperature. The heat lost by the metal is $(210)(0.250)(85 - T_f)$, and the heat lost by the calorimeter is $(345)(0.100)(85 - T_f)$. The heat gained by the water is $(95)(1.00)(T_f - 21.3)$.Setting the heat lost equal to the heat gained and solving for $T_f$ gives us the final temperature.

Question 17

Find the final equilibrium temperature when 20.0 g of milk at 15.0°C is added to 155.0 g of coffee at 89.0°C.(Assume the specific heats of coffee and milk are the same as water and neglect the heat capacity of the container. )c_water = 1.00 cal/g⋅°C = 4 186 J/kg⋅°C A)80.5°C B)100°C C)104°C D)77.1°C E)78.9°C

Accepted Answer

The correct answer is A because the final temperature will be between the initial temperatures of the coffee and the milk. The specific heat of water is 1.00 cal/g⋅°C, so the heat lost by the coffee will be equal to the heat gained by the milk. The final temperature will be the average of the two initial temperatures, which is 80.5°C.

Question 18

Pennies used to be made of copper,but now they are made of copper-coated zinc.If one were to do a precise calorimetry experiment to determine the specific heat of the new pennies,what would the result be?
A)It would be that of copper since copper is on the outside.
B)It would be that of zinc since zinc is in the center.
C)It would be the sum of the copper and zinc specific heats.
D)It would be between that of copper and that of zinc,depending on coating thickness.

Accepted Answer

The specific heat of the penny will be between that of copper and zinc because the coating thickness can vary. The specific heat of the penny will depend on the proportion of copper and zinc in the coin and the thickness of the coating. Therefore, it would not be the specific heat of just copper or zinc or the sum of the two.

Question 19

A 140-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 290 g of water.The water temperature rises from 6.8°C to 26.8°C.Given c_Cu = 0.10 cal/g⋅°C,and c_water = 1.00 cal/g⋅°C,what was the temperature of the kiln? A)720°C B)387°C C)682°C D)441°C E)68.2°C

Accepted Answer

The answer of A 140-g block of copper is taken...

Question 20

A plot of the temperature versus the energy per kg added to a piece of ice as it goes from below freezing at −10°C to becoming steam at 110°C consists of straight lines,some horizontal and some with an upward slope.What do the upward slopes represent?
A)specific heats
B)reciprocals of specific heats
C)latent heats
D)reciprocals of latent heats

Accepted Answer

The answer of A plot of the temperature versus the...

Quiz 11: Energy in Thermal Processes

In typical calorimetry problems,what is the usual conservation law being used?

A swimming pool heater has to be able to raise the temperature of the 50 000 gallons of water in the pool by 12.0 C°.How many kilowatt-hours of energy are required? (One gallon of water has a mass of approximately 3.8 kg and the specific heat of water is 4,186 J/kg⋅°C. )

A 3.00-kg copper rod is 45.0 cm long at 24.8°C.If 45 000 J are transferred to the rod by heat,what is its change in length? c_copper = 387 J/kg⋅°C and α_copper = 17.0 × 10−⁶/°C.

Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C.Thermal equilibrium is reached at 30°C.The specific heat of the solid

Which of the following has the highest specific heat?

What is the temperature increase of 3.00 kg of water when heated by a 900-W immersion heater for 9.00 min? (c_w = 4 186 J/kg⋅°C)

A slice of bread contains about 125 kcal.If specific heat of a person were 1.00 kcal/kg⋅°C,by how many °C would the temperature of a 65.0-kg person increase if all the energy in the bread were converted to heat?

A piece of copper of mass 110 g is being drilled through with a ¹/₂" electric drill.The drill operates at 25.0 W and takes 25.0 s to bore through the copper.If all the energy from the drill heats the copper,find the copper's increase in temperature.c_copper = 387 J/kg⋅°C.

A solar heating system has a 25.0% conversion efficiency;the solar radiation incident on the panels is 800 W/m².What is the increase in temperature of 25.0 kg of water in a 1.00-h period by a 3.40-m²-area collector? (c_w = 4 186 J/kg⋅°C)

A 4.00-g lead bullet is traveling at a speed of 200 m/s when it embeds in a wood post.If we assume that half of the resultant heat energy generated remains with the bullet,what is the increase in temperature of the embedded bullet? (specific heat of lead = 0.0305 kcal/kg⋅°C,1 kcal = 4 186 J)

A machine gear consists of 0.14 kg of iron and 0.19 kg of copper.How much total heat is generated in the part if its temperature increases by 35.0 C°? (Specific heats of iron and copper are 450 and 390 J/kg⋅°C,respectively. )

A waterfall is 128 m high.What is the increase in water temperature at the bottom of the falls if all the initial potential energy goes into heating the water? (g = 9.8 m/s²,c_w = 4 186 J/kg⋅°C)

Find the final equilibrium temperature when 20.0 g of milk at 15.0°C is added to 155.0 g of coffee at 89.0°C.(Assume the specific heats of coffee and milk are the same as water and neglect the heat capacity of the container. ) c_water = 1.00 cal/g⋅°C = 4 186 J/kg⋅°C

Pennies used to be made of copper,but now they are made of copper-coated zinc.If one were to do a precise calorimetry experiment to determine the specific heat of the new pennies,what would the result be?

A 140-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 290 g of water.The water temperature rises from 6.8°C to 26.8°C.Given c_Cu = 0.10 cal/g⋅°C,and c_water = 1.00 cal/g⋅°C,what was the temperature of the kiln?

A plot of the temperature versus the energy per kg added to a piece of ice as it goes from below freezing at −10°C to becoming steam at 110°C consists of straight lines,some horizontal and some with an upward slope.What do the upward slopes represent?