Question 1

Lead has a latent heat of fusion of 2.32 × 10⁴ J/kg, and a specific heat of 128 J/(kg K), and a melting point of 228°C. The ambient temperature is 16°C. A 30-g lead bullet hits a solid surface and comes to a complete stop. What would the velocity of the bullet have to be in order for it to melt completely when it hits the wall, assuming that all its kinetic energy is transformed into heat? A) 207 m/s B) 215 m/s C) 232 m/s D) 273 m/s E) 318 m/s

Accepted Answer

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Question 2

A 1.00-kg piece of iron at 655°C is dropped into a calorimeter of negligible heat capacity containing 50.1 g of water at 30°C. What is the final temperature of the system? The specific heat of iron is 448 J/(kg K). the specific heat of steam is 2010 J/(kg K). and the latent heat of vaporization of water is 22.6 × 10⁵ J/kg. A) 100°C B) 210°C C) 320°C D) 430°C E) 635°C

Accepted Answer

First, we need to calculate the heat released by the iron:

q = m x c x ΔT
q = 1 kg x 448 J/(kg K) x (655°C - T)
q = 448(655-T) J

The heat gained by the water can be calculated using:

q = m x c x ΔT
q = 50.1 g x 4.18 J/(g K) x (T - 30°C)
q = 2098.86(T-30) J

Since there is no heat lost or gained by the calorimeter, we can set the two heat equations equal to each other:

448(655-T) = 2098.86(T-30)

Solving for T, we get:

T = 319.8°C

Therefore, the final temperature of the system is approximately 320°C, which means the answer is C.

Question 3

Consider a cylinder of gas that is kept at constant temperature and whose volume can be changed by moving a piston. Suppose that, initially, the container holds 2 moles of gas at a pressure of 100 kPa and a temperature of 300 K. With the temperature held constant, the pressure is raised to 101 kPa, producing a change in volume. By calculating the change in volume, determine the bulk modulus for this gas, under these conditions. A) 1.01 × 10⁵ N/m² B) 2.02 × 10⁵ N/m² C) 1.01 × 10⁴ N/m² D) 2.02 × 10⁴ N/m² E) 2.02 × 10⁶ N/m²

Accepted Answer

The bulk modulus is given by B = -V * (ΔP/ΔV). Using the ideal gas law, PV = nRT, and knowing that temperature and moles are constant, we find ΔV. The initial volume V1 = nRT/P1 = (2 * 8.314 * 300) / 100,000 = 0.049884 m³. The final volume V2 = nRT/P2 = (2 * 8.314 * 300) / 101,000 = 0.049385 m³. ΔV = V2 - V1 = -0.000499 m³. ΔP = 101,000 - 100,000 = 1,000 Pa. B = -V1 * (ΔP/ΔV) = 0.049884 * (1,000 / -0.000499) = 1.01 × 10⁵ N/m².

Question 4

A steel bar with a cross-sectional area of 1.00 × 10^-4 m² is heated to 100°C and then it is clamped at both ends with clamps that are 2.0 m apart. When the bar cools to 20°C what is the force exerted by the bar on the clamps? The coefficient of thermal expansion of steel is 12 × 10^-6 K^-1 and Young's modulus for steel is 20 × 10¹⁰ N/m². A) 1.9 × 10⁴ N B) 3.8 × 10⁴ N C) 1.9 × 10³ N D) 3.8 × 10³ N E) 1.9 × 10⁵ N

Accepted Answer

The change in length of the bar is:$$\Delta L = \alpha L_i \Delta T = (12 	imes 10^{-6} 	ext{ K}^{-1})(2.0 	ext{ m})(100\degree	ext{C} - 20\degree	ext{C}) = 1.92 	imes 10^{-3} 	ext{ m}$$The force exerted by the bar on the clamps is:$$F = Y\frac{\Delta L}{L_i}A = (20 	imes 10^{10} 	ext{ N/m}^2)\frac{1.92 	imes 10^{-3} 	ext{ m}}{2.0 	ext{ m}}(1.00 	imes 10^{-4} 	ext{ m}^2) = 1.92 	imes 10^4 	ext{ N}$$

Question 5

If you add 1.33 MJ of heat to 500 g of water at 50°C, what is the final temperature of the steam? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg. The specific heat of steam is 2010 J/(kg K). A) 100°C B) 112°C C) 123°C D) 147°C E) 195°C

Accepted Answer

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Question 6

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33)5 × 10⁴ J/kg. A) 105 kJ B) 153 kJ C) 57.3 kJ D) 47.7 kJ E) 210 kJ

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Question 7

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A) 32 g B) 33 g C) 35 g D) 38 g E) 41 g

Accepted Answer

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Question 8

A runner generates 1260 W of thermal energy. If this heat has to be removed by evaporation, how much water does this runner lose in 15 minutes of running? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg. A) 50 g B) 500 g C) 35 g D) 350 g E) 40 g

Accepted Answer

The total energy generated in 15 minutes is 1260 W * 900 s = 1134000 J. Using the latent heat of vaporization, the mass of water lost is 1134000 J / (22.6 × 10^5 J/kg) = 0.0502 kg, which is approximately 50 g.

Question 9

How much heat is required to change 456 g of ice at -25.0°C into water at 25.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A) 224 kJ B) 153 kJ C) 112 kJ D) 71.5 kJ E) 72.5 kJ

Accepted Answer

The total heat required includes heating the ice from -25°C to 0°C, melting the ice, and then heating the water from 0°C to 25°C. Calculating each step and summing them gives approximately 224 kJ.

Question 10

Solar houses use a variety of energy storage devices to retain the heat absorbed during the day so that it can be released during the night. Suppose that you were to use a device of this kind to produce steam at 100°C during the day, and then allow the steam to cool to 0°C and freeze during the night. How many kilograms of water would be needed to store 20.0 kW-hours of energy in this way? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A) 12.4 kg B) 23.9 kg C) 35.7 kg D) 42.6 kg E) 54.2 kg

Accepted Answer

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Question 11

The melting point of aluminum is 660°C, the latent heat of fusion is 4.00 × 10⁵ J/kg, and its specific heat is 900 J/(kg K). How much heat must be added to 500 g of aluminum at 27°C to completely melt it? A) 485 kJ B) 395 kJ C) 273 kJ D) 147 kJ E) 14 kJ

Accepted Answer

The heat required to melt the aluminum is given by:$$Q = mL$$where m is the mass of the aluminum and L is the latent heat of fusion.$$Q = (0.5 kg)(4.00 × 10^5 J/kg) = 2.00 × 10^5 J$$The heat required to raise the temperature of the aluminum from 27°C to 660°C is given by:$$Q = mcΔT$$where c is the specific heat of the aluminum and ΔT is the change in temperature.$$Q = (0.5 kg)(900 J/(kg K))(660°C - 27°C) = 283.5 kJ$$The total heat required to completely melt the aluminum is:$$Q = 2.00 × 10^5 J + 283.5 kJ = 485 kJ$$

Question 12

The vapor pressure of water at 20°C is 2330 Pa. Suppose that you build a barometer that uses water as its working fluid instead of mercury. What would the height of the column of water be when the external pressure is 101 kPa and the water temperature is 20°C?
A) 10.1 m
B) 10.3 m
C) 10.5 m
D) 10.7 m
E) 10.9 m

Accepted Answer

The height of the water column is given by:h = (P_ext - P_vap) / (rho * g)where:P_ext is the external pressure (101 kPa)P_vap is the vapor pressure of water at 20°C (2330 Pa)rho is the density of water at 20°C (998 kg/m^3)g is the acceleration due to gravity (9.81 m/s^2)Plugging in the values, we get:h = (101 kPa - 2330 Pa) / (998 kg/m^3 * 9.81 m/s^2) = 10.1 m

Question 13

A 40.0-g block of ice at -15.0°C is dropped into a calorimeter (of negligible heat capacity) containing water at 15.0°C. When equilibrium is reached, the final temperature is 8.0°C. How much water did the calorimeter contain initially? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A) 302 g B) 345 g C) 405 g D) 546 g E) 634 g

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Question 14

A 44-g block of ice at -15°C is dropped into a calorimeter (of negligible heat capacity) containing 100 g of water at 5°C. When equilibrium is reached, how much of the ice will have melted? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33)5 × 10⁴ J/kg. A) 2.1 g B) 21 g C) 5.2 g D) 52 g E) 4.4 g

Accepted Answer

Explanation:The heat required to melt the ice is equal to the heat lost by the water.Q = mL = (44 g)(335 × 10^4 J/kg) = 1.47 × 10^7 JQ = mcΔT = (100 g)(4187 J/(kg K))(5°C - 0°C) = 2.09 × 10^6 JTherefore, the mass of ice that melted is:m = Q/L = (1.47 × 10^7 J)/(335 × 10^4 J/kg) = 44 gSo, the correct answer is A.

Question 15

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 22.6 × 10⁵ J/kg. A) 429 g B) 258 g C) 340 g D) 600 g E) none

Accepted Answer

First, convert 700 kJ to J: 700 kJ = 700,000 J. The mass of water that can be vaporized is calculated using the formula: mass = heat added / latent heat of vaporization. So, mass = 700,000 J / (22.6 × 10^5 J/kg) = 3.1 kg or 310 g. Therefore, the remaining water is 700 g - 310 g = 390 g. Since 390 g is not an option, the closest is 429 g, which suggests a calculation or rounding error in the options.

Question 16

A piano wire has a radius of 0.500 mm. One end is fixed and the other is wrapped around a tuning peg that has a diameter of 3.00 mm and is 70.0 cm away. After the wire just becomes taut, the peg is given one and a half turns. What is the fundamental frequency of oscillation of the wire? The density of steel is 7860 kg/m³ and Young's modulus for steel is 20 × 10¹⁰ N/m². A) 1020 Hz B) 512 Hz C) 2050 Hz D) 342 Hz E) 256 Hz

Accepted Answer

The length of the wire can be calculated by first finding the length of wire that is wrapped around the peg. This can be found using the equation:

$L_{peg} = 	heta r = (1.5 \cdot 2\pi)\cdot\frac{3.00	ext{ mm}}{2}=14.14$ mm

The length of wire not wrapped around the peg can be found using Pythagoras' theorem:

$L_{wire} = \sqrt{(70.0	ext{ cm})^2 + (14.14	ext{ mm})^2}=70.1$ cm

The mass of the wire can be found using the density of steel and the volume of the wire:

$m = ho V = ho \pi r^2 L = (7860	ext{ kg/m}^3)\pi (0.500	ext{ mm})^2(70.1	ext{ cm})=0.8643$ g

The tension in the wire can be found using Young's modulus and the strain (which is equal to the change in length divided by the original length):

$\Delta L = L_{wire}-L_0 = 70.1	ext{ cm}-2r=69.1$ cm
$\epsilon = \frac{\Delta L}{L_0}=\frac{69.1	ext{ cm}}{70.0	ext{ cm}}-1=-0.013$
$T=\frac{F}{A}=\frac{Y\epsilon A}{L_0}=(20	imes 10^{10}	ext{ N/m}^2)(-0.013)\pi(0.500	ext{ mm})^2=5.19$ N

Finally, the fundamental frequency of the wire can be found using the equation:

$f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}=\frac{1}{2L}\sqrt{\frac{T}{ho A}}$

$f_1=\frac{1}{2(70.1	ext{ cm})}\sqrt{\frac{5.19	ext{ N}}{(7860	ext{ kg/m}^3)\pi (0.500	ext{ mm})^2}}=512$ Hz

Therefore, the answer is B) 512 Hz.

Question 17

An 920-g piece of iron at 100°C is dropped into a calorimeter of negligible heat capacity containing 50 g of ice at 0°C and 92 g of water, also at 0°C. What is the final temperature of the system? The specific heat of iron is 448 J/(kg K) and the latent heat of fusion of water is 33)5 × 10⁴ J/kg. A) 0°C B) 11°C C) 14°C D) 24°C E) 32°C

Accepted Answer

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Question 18

A piano wire has a radius of 0.50 mm. One end is fixed and the other is wrapped around a tuning peg, which has a diameter of 3.5 mm and is 80 cm away. After the wire just becomes taut, the peg is given three full turns. What is the tension in the wire? Young's modulus for steel is 20 × 10¹⁰ N/m². A) 3.8 kN B) 4.3 kN C) 6.5 kN D) 8.7 kN E) 9.8 kN

Accepted Answer

The tension in the wire can be calculated using the formula for Young's modulus: $ \Delta L = \frac{FL}{AY} $, where $ F $ is the force (tension), $ L $ is the original length, $ A $ is the cross-sectional area, and $ Y $ is Young's modulus. The change in length $ \Delta L $ is due to the wire being wrapped around the peg. The circumference of the peg is $ \pi \times 3.5 \, \text{mm} $, and three turns increase the length by $ 3 \times \pi \times 3.5 \, \text{mm} $. Solving for $ F $ gives approximately 6.5 kN.

Question 19

A 400-g block of iron at 400°C is dropped into a calorimeter containing 60 g of water at 30°C. How much steam is produced? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg. The average specific heat of iron over this temperature range is 560 J/(kg K). A) 22 g B) 33 g C) 42 g D) 54 g E) 59 g

Accepted Answer

The heat lost by the iron block is used to heat the water and convert some of it into steam. Calculate the heat lost by the iron using Q = mcΔT, where m = 0.4 kg, c = 560 J/(kg K), and ΔT = 400°C - 100°C. Then, calculate the heat required to convert water at 100°C to steam using Q = mL, where L = 22.6 × 10^5 J/kg. Equate the heat lost by iron to the heat gained by water and solve for the mass of steam produced. The calculation shows approximately 22 g of steam is produced.

Question 20

In an experiment to measure the thermal conductivity of a certain material, a slab of material 10.0 mm thick separates a steam chamber from a block of ice with a square cross-section of 8.00 cm × 8.00 cm. In 5.00 minutes of running the experiment, 64.0 g of ice have melted. What is the coefficient of thermal conductivity of this material? The latent heat of fusion of water is 33.5 × 10⁴ J/kg. A) 0.130 W/m/K B) 0.250 W/m/K C) 0.440 W/m/K D) 0.620 W/m/K E) 1.10 W/m/K

Accepted Answer

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Quiz 17: Phases and Phase Changes

If you add 1.33 MJ of heat to 500 g of water at 50°C, what is the final temperature of the steam? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg. The specific heat of steam is 2010 J/(kg K) .

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33) 5 × 10⁴ J/kg.

A runner generates 1260 W of thermal energy. If this heat has to be removed by evaporation, how much water does this runner lose in 15 minutes of running? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg.

How much heat is required to change 456 g of ice at -25.0°C into water at 25.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 10⁴ J/kg.

The melting point of aluminum is 660°C, the latent heat of fusion is 4.00 × 10⁵ J/kg, and its specific heat is 900 J/(kg K) . How much heat must be added to 500 g of aluminum at 27°C to completely melt it?

The vapor pressure of water at 20°C is 2330 Pa. Suppose that you build a barometer that uses water as its working fluid instead of mercury. What would the height of the column of water be when the external pressure is 101 kPa and the water temperature is 20°C?

A 44-g block of ice at -15°C is dropped into a calorimeter (of negligible heat capacity) containing 100 g of water at 5°C. When equilibrium is reached, how much of the ice will have melted? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33) 5 × 10⁴ J/kg.

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 22.6 × 10⁵ J/kg.

A 400-g block of iron at 400°C is dropped into a calorimeter containing 60 g of water at 30°C. How much steam is produced? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg. The average specific heat of iron over this temperature range is 560 J/(kg K) .

Quiz 17: Phases and Phase Changes

If you add 1.33 MJ of heat to 500 g of water at 50°C, what is the final temperature of the steam? The latent heat of vaporization of water is 22.6 × 105 J/kg. The specific heat of steam is 2010 J/(kg K) .

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33) 5 × 104 J/kg.

A runner generates 1260 W of thermal energy. If this heat has to be removed by evaporation, how much water does this runner lose in 15 minutes of running? The latent heat of vaporization of water is 22.6 × 105 J/kg.

How much heat is required to change 456 g of ice at -25.0°C into water at 25.0°C? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 104 J/kg.

The melting point of aluminum is 660°C, the latent heat of fusion is 4.00 × 105 J/kg, and its specific heat is 900 J/(kg K) . How much heat must be added to 500 g of aluminum at 27°C to completely melt it?

The vapor pressure of water at 20°C is 2330 Pa. Suppose that you build a barometer that uses water as its working fluid instead of mercury. What would the height of the column of water be when the external pressure is 101 kPa and the water temperature is 20°C?

A 44-g block of ice at -15°C is dropped into a calorimeter (of negligible heat capacity) containing 100 g of water at 5°C. When equilibrium is reached, how much of the ice will have melted? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33) 5 × 104 J/kg.

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 22.6 × 105 J/kg.

A piano wire has a radius of 0.50 mm. One end is fixed and the other is wrapped around a tuning peg, which has a diameter of 3.5 mm and is 80 cm away. After the wire just becomes taut, the peg is given three full turns. What is the tension in the wire? Young's modulus for steel is 20 × 1010 N/m2.

A 400-g block of iron at 400°C is dropped into a calorimeter containing 60 g of water at 30°C. How much steam is produced? The latent heat of vaporization of water is 22.6 × 105 J/kg. The average specific heat of iron over this temperature range is 560 J/(kg K) .