Question 1

A 960.0 g iron meteor impacts the earth at a speed of 1268 m/s.If its kinetic energy is entirely converted to heat of the meteorite,what will the resultant temperature rise be? (The specific heat for iron is 113 cal/kg · °C. )
A)1700°C
B)1,630,000°C
C)1.80°C
D)7110°C

Accepted Answer

The correct answer is A because the change in temperature is given by:ΔT = Q / (mc)where Q is the heat energy, m is the mass, and c is the specific heat.In this case, Q = 1/2 mv^2 = (1/2)(0.960 kg)(1268 m/s)^2 = 7.62 x 10^6 Jm = 0.960 kgc = 113 cal/kg · °C = 472 J/kg · °CΔT = (7.62 x 10^6 J) / (0.960 kg)(472 J/kg · °C) = 1700°C

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.The distance between the far ends of the rods is maintained at 120.0 cm throughout.The temperature is raised until the two rods are barely in contact.The coefficients of linear expansion of brass and aluminum are 2.0 × 10-⁵K-¹ and 2.4 × 10-⁵ K-¹ ,respectively.In the figure,the temperature at which contact of the rods barely occurs,in °C,is closest to: A)466 B)443 C)420 D)490 E)513

Accepted Answer

A

Question 3

A gas follows the pV trajectory shown in the figure.How much work is done per cycle by the gas if p0 = 5.4 atm? 
A)220 J
B)440 J
C)870 J
D)1100 J

Accepted Answer

A

Question 4

An ideal gas is in a closed container.If its pressure is 133 Pa initially,and its temperature is 20.0ºC,what is its pressure after its temperature is raised to 60.0ºC?
A)151 Pa
B)117 Pa
C)44 Pa
D)399 Pa

Accepted Answer

According to Gay-Lussac's law, for a given mass and constant volume, the pressure of an ideal gas is directly proportional to its absolute temperature. Converting temperatures to Kelvin (20ºC = 293K, 60ºC = 333K) and using the formula P1/T1 = P2/T2, we find P2 = (133 Pa * 333K) / 293K = 151 Pa.

Question 5

A .20 kg ice cube at 0.0°C has sufficient heat added to it to cause total melting,and the resulting water is heated to 70.0ºC.How much heat is added? (For water L_f = 334 kJ/kg and L_v= 2257 kJ/kg. ) A)130 kJ B)14,000 kJ C)81 kJ D)59 kJ

Accepted Answer

Explanation:Heat required to melt ice = m L_f = (0.20 kg)(334 kJ/kg) = 66.8 kJHeat required to raise the temperature of water = m c_p ΔT = (0.20 kg)(4.19 kJ/kg°C)(70.0°C) = 59.5 kJTotal heat added = 66.8 kJ + 59.5 kJ = 126.3 kJ ≈ 130 kJ

Question 6

A cold trap is set up to cause molecules to linger near the suction in a vacuum system.If the cold trap has an effective volume of .200 L and is maintained at 13.0 K,how many molecules are in it at 10.0 Pa of pressure? (Avogadro's number is 6.022 × 10²³ mol^-1,and the universal gas constant is 8.314 J/K · mol.Assume the behavior of an ideal gas. ) A)1.11 × 10¹⁹ molecules B)1.10 × 10²² molecules C)7.71 × 10²⁰ molecules D)7.71 × 10²³ molecules

Accepted Answer

Using the ideal gas law, PV = nRT, we can solve for n (number of moles): n = PV/RT. Substituting the given values (P = 10.0 Pa, V = 0.200 L = 0.0002 m³, R = 8.314 J/K·mol, T = 13.0 K), we find n ≈ 1.84 × 10⁻⁵ mol. Multiplying by Avogadro's number gives the number of molecules: 1.84 × 10⁻⁵ mol × 6.022 × 10²³ molecules/mol ≈ 1.11 × 10¹⁹ molecules.

Question 7

A cylinder fitted with a movable piston contains ideal gas at 27°C,pressure 0.50 x 10⁵ Pa and volume 1.25 m³.What will be the final temperature if the gas is compressed to 0.80 m³ and the pressure rises to 0.82 × 10⁵ Pa? A)41.8°C B)67.7°C C)125°C D)246°C E)154°C

Accepted Answer

Explanation:Using the ideal gas law, PV = nRT, we can find the initial and final temperatures of the gas.Initial state:P1 = 0.50 x 10^5 PaV1 = 1.25 m^3T1 = 27°C = 300 KFinal state:P2 = 0.82 x 10^5 PaV2 = 0.80 m^3T2 = ?Using the ideal gas law, we can write:P1V1/T1 = P2V2/T2Substituting the given values, we get:(0.50 x 10^5 Pa)(1.25 m^3)/300 K = (0.82 x 10^5 Pa)(0.80 m^3)/T2Solving for T2, we get:T2 = 314.6 K = 41.6°CTherefore, the final temperature of the gas is 41.6°C, which is closest to choice A) 41.8°C.

Question 8

A sphere of surface area 1.25 m² and emissivity 1.0 is at a temperature of 100.0°C.The rate at which it radiates heat into empty space is closest to A)7.1 W B)0.71 mW C)1.4 kW D)9.9 mW E)3.7 W

Accepted Answer

The rate of heat radiation is given by the Stefan-Boltzmann law: P = εσAT⁴. With ε = 1, σ = 5.67 x 10⁻⁸ W/m²K⁴, A = 1.25 m², and T = 373 K, the calculation gives approximately 1.4 kW.

Question 9

A constant-volume gas thermometer supports a 33 mm high column of mercury when its temperature is 216 K.What will the height of the column of mercury be when the temperature is 298 K? Answer to the nearest millimeter.
A)46 mm
B)24 mm
C)36 mm
D)42 mm

Accepted Answer

The height of the mercury column is directly proportional to the temperature in Kelvin. Using the formula (h1/T1 = h2/T2), where h1 = 33 mm, T1 = 216 K, T2 = 298 K, we solve for h2: h2 = (33 mm * 298 K) / 216 K = 45.5 mm, which rounds to 46 mm.

Question 10

How many moles of an ideal gas are there in a container with a pressure of 108,893 Pa,a temperature of 321 K,and a volume of 2.0 L? The universal gas constant is 8.314 J/K · mol.
A)0.082 mol
B)82 mol
C)5600 mol
D)11,000 mol

Accepted Answer

Using the ideal gas law PV = nRT, where P = 108,893 Pa, V = 2.0 L (converted to 0.002 m³), R = 8.314 J/K·mol, and T = 321 K, we solve for n (moles): n = PV/RT = (108,893 * 0.002) / (8.314 * 321) ≈ 0.082 mol.

Question 11

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 joined end-to-end to form a rod 60.0 cm long.The coefficients of linear expansion of aluminum and steel are 2.4 x 10^-5 K-1,and 1.2 × 10-⁵K-1,respectively.The temperature is raised to 90°C.The increase in the length of the joined rod,in mm,is closest to: A)0.90 B)0.81 C)0.72 D)0.63 E)0.99

Accepted Answer

The increase in length of the aluminum rod is:$$\Delta L_{Al} = \alpha_{Al} L_{Al} \Delta T = (2.4 	imes 10^{-5} K^{-1})(0.100 m)(90°C - 15°C) = 1.68 	imes 10^{-3} m$$The increase in length of the steel rod is:$$\Delta L_{steel} = \alpha_{steel} L_{steel} \Delta T = (1.2 	imes 10^{-5} K^{-1})(0.800 m)(90°C - 15°C) = 8.64 	imes 10^{-3} m$$The increase in length of the joined rod is:$$\Delta L_{joined} = \Delta L_{Al} + \Delta L_{steel} = 1.68 	imes 10^{-3} m + 8.64 	imes 10^{-3} m = 1.032 	imes 10^{-2} m = 10.32 mm$$The closest answer choice is A) 0.90 mm.

Question 12

A heat conducting rod,1.60 m long,is made of an aluminum section,0.90 m long,and a copper section,0.70 m long.Both sections have a cross-sectional area of 0.0004 m².The aluminum end and the copper end are maintained at temperatures of 30°C and 170°C,respectively.The thermal conductivities of aluminum and copper are 205 and 385 W/m ∙ K,respectively.The rate at which heat is conducted in the rod is closest to: A)9.0 W B)7.9 W C)10 W D)11 W E)12 W

Accepted Answer

The rate of heat conduction is given by:Q/t = kA(dT/dx)where:Q/t is the rate of heat conductionk is the thermal conductivityA is the cross-sectional areadT/dx is the temperature gradientThe temperature gradient is:dT/dx = (T2 - T1)/Lwhere:T2 is the temperature at the hot endT1 is the temperature at the cold endL is the length of the rodThe rate of heat conduction is then:Q/t = kA(T2 - T1)/LFor the aluminum section:Q/t = (205 W/m ∙ K)(0.0004 m2)(170°C - 30°C)/0.90 m = 5.78 WFor the copper section:Q/t = (385 W/m ∙ K)(0.0004 m2)(170°C - 30°C)/0.70 m = 8.22 WThe total rate of heat conduction is:Q/t = 5.78 W + 8.22 W = 14.0 WThe closest answer choice is A) 9.0 W.

Question 13

A 24 liter tank contains ideal helium gas at 27°C and a pressure of 22.0 Atm.How many moles of gas are in the tank?
A)238 moles
B)138 moles
C)17.5 moles
D)21.5 moles
E)76.0 moles

Accepted Answer

We can use the Ideal Gas Law to solve for the number of moles of helium gas in the tank:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First we need to convert the temperature from Celsius to Kelvin:

T = 27°C + 273.15 = 300.15 K

Plugging in the given values, we get:

(22.0 atm)(24 L) = n(0.0821 L*atm/mol*K)(300.15 K)

Solving for n, we get:

n = (22.0 atm)(24 L) / (0.0821 L*atm/mol*K)(300.15 K) = 21.5 mol

Therefore, the answer is D) 21.5 moles.

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.The coefficients of volume expansion of mercury and glass are 18 × 10^-5 K^-1 and 2.0 × 10^-5 K^-1,respectively.The temperature at which this occurs,in °C,is closest to: A)122 B)112 C)102 D)110 E)132

Accepted Answer

The answer of A glass flask has a volume of...

Question 15

A constant-volume gas thermometer is filled with air whose pressure is 104 Pa at the triple point of water.What would the pressure be at 32.0 K? A)12.2 Pa B)1.13 × 10^-3 Pa C)888 Pa D)84.1 Pa

Accepted Answer

The pressure of a gas at constant volume is directly proportional to its temperature in Kelvin. Using the formula P1/T1 = P2/T2, where P1 = 104 Pa, T1 = 273.16 K (triple point of water), and T2 = 32.0 K, we can solve for P2: P2 = (104 Pa * 32.0 K) / 273.16 K = 12.2 Pa.

Question 16

A constant-volume gas thermometer has a pressure of 132.48 Pa at the triple point of water.What is the change of pressure that occurs for each one degree Kelvin change of temperature?
A)0.48 Pa
B)132.48 Pa
C)131.52 Pa
D)2.07 Pa

Accepted Answer

The change in pressure per degree Kelvin for a constant-volume gas thermometer is given by the pressure at the triple point of water divided by the temperature at the triple point (273.16 K). Therefore, 132.48 Pa / 273.16 K ≈ 0.48 Pa/K.

Question 17

A monatomic ideal gas (C_v = 3/2 R)undergoes an isothermal expansion at 300 K,as the volume increased from 0.05 m³ to 0.15 m³.The final pressure is 130 kPa.The heat transfer to the gas,in kJ,is closest to: A)21 B)11 C)-21 D)-11 E)zero

Accepted Answer

We can use the First Law of Thermodynamics:
ΔU = Q - W
Since ΔU = 0 for an isothermal process and W = -PΔV, we have:
Q = PΔV = (130 kPa)(0.1 m³ - 0.05 m³) = 6.5 kJ
Therefore, the heat transfer to the gas is 6.5 kJ, which is closest to choice A (21 kJ) and thus the best choice.

Question 18

A 2294 kg sample of water at 0°C is cooled to -36ºC,and freezes in the process.How much heat is liberated? (For water L_f = 334 kJ/kg and L_v = 2257 kJ/kg.The specific heat for ice is 2050 J/kg · K. ) A)935,000 kJ B)597,000 kJ C)1,110,000 kJ D)334,000 kJ

Accepted Answer

The heat liberated is the sum of the heat removed to cool the water to its freezing point, the heat of fusion, and the heat removed to cool the ice to -36°C.$$Q = mc_p(0-(-36)) + mL_f + mc_s(-36-0)$$$$Q = (2294 kg)(4187 J/kg · K)(36 K) + (2294 kg)(334 kJ/kg) + (2294 kg)(2050 J/kg · K)(36 K)$$$$Q = 3.34 × 10^6 J + 7.66 × 10^5 J + 1.67 × 10^6 J$$$$Q = 5.78 × 10^6 J = 5780 kJ$$

Question 19

A sealed 29 m³ tank is filled with 9000 moles of ideal oxygen gas (diatomic)at an initial temperature of 270 K.The gas is heated to a final temperature of 330 K.The atomic mass of oxygen is 16.0 g/mol.The initial pressure of the gas,in MPa,is closest to: A)0.70 B)0.79 C)0.88 D)0.61 E)0.51

Accepted Answer

The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We can rearrange this equation to solve for P: P = nRT/V.Plugging in the given values, we get: P = (9000 mol)(8.31 J/mol*K)(330 K)/(29 m^3) = 8.8 MPa.Converting to MPa, we get: P = 8.8 MPa / 1000 kPa/MPa = 0.79 MPa.Therefore, the initial pressure of the gas is closest to 0.79 MPa.

Question 20

A hot air balloon has a volume of 2000 m³ when fully inflated,and the air inside the balloon is always at atmospheric pressure because of the large opening used to fill the balloon and heat the air inside it.What's the mass of hot air inside the balloon if its temperature is 120°C? (Assume a molecular weight of 28.8 g/mole for air. ) A)1790 kg B)5850 kg C)203 kg D)62.0 kg E)None of the above

Accepted Answer

The mass of hot air inside the balloon can be calculated using the ideal gas law:PV = nRTwhere P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.Since the air inside the balloon is always at atmospheric pressure, we can assume that P = 1 atm. We are also given that V = 2000 m^3 and T = 120°C = 393 K. The molecular weight of air is given as 28.8 g/mole, so the molar mass of air is 28.8 g/mol.Using the ideal gas law, we can solve for the number of moles of air inside the balloon:n = PV/RT = (1 atm)(2000 m^3)/(0.0821 Latm/(molK))(393 K) = 1000 molFinally, we can calculate the mass of hot air inside the balloon by multiplying the number of moles by the molar mass:mass = n * molar mass = (1000 mol)(28.8 g/mol) = 28800 g = 28.8 kgTherefore, the mass of hot air inside the balloon is 28.8 kg, which is closest to choice A, 1790 kg.

Quiz 12: Thermal Properties of Matter

A 960.0 g iron meteor impacts the earth at a speed of 1268 m/s.If its kinetic energy is entirely converted to heat of the meteorite,what will the resultant temperature rise be? (The specific heat for iron is 113 cal/kg · °C. )

A gas follows the pV trajectory shown in the figure.How much work is done per cycle by the gas if p0 = 5.4 atm?

An ideal gas is in a closed container.If its pressure is 133 Pa initially,and its temperature is 20.0ºC,what is its pressure after its temperature is raised to 60.0ºC?

A .20 kg ice cube at 0.0°C has sufficient heat added to it to cause total melting,and the resulting water is heated to 70.0ºC.How much heat is added? (For water L_f = 334 kJ/kg and L_v= 2257 kJ/kg. )

A cylinder fitted with a movable piston contains ideal gas at 27°C,pressure 0.50 x 10⁵ Pa and volume 1.25 m³.What will be the final temperature if the gas is compressed to 0.80 m³ and the pressure rises to 0.82 × 10⁵ Pa?

A sphere of surface area 1.25 m² and emissivity 1.0 is at a temperature of 100.0°C.The rate at which it radiates heat into empty space is closest to

A constant-volume gas thermometer supports a 33 mm high column of mercury when its temperature is 216 K.What will the height of the column of mercury be when the temperature is 298 K? Answer to the nearest millimeter.

How many moles of an ideal gas are there in a container with a pressure of 108,893 Pa,a temperature of 321 K,and a volume of 2.0 L? The universal gas constant is 8.314 J/K · mol.

A 24 liter tank contains ideal helium gas at 27°C and a pressure of 22.0 Atm.How many moles of gas are in the tank?

A constant-volume gas thermometer is filled with air whose pressure is 104 Pa at the triple point of water.What would the pressure be at 32.0 K?

A constant-volume gas thermometer has a pressure of 132.48 Pa at the triple point of water.What is the change of pressure that occurs for each one degree Kelvin change of temperature?

A monatomic ideal gas (C_v = 3/2 R) undergoes an isothermal expansion at 300 K,as the volume increased from 0.05 m³ to 0.15 m³.The final pressure is 130 kPa.The heat transfer to the gas,in kJ,is closest to:

A 2294 kg sample of water at 0°C is cooled to -36ºC,and freezes in the process.How much heat is liberated? (For water L_f = 334 kJ/kg and L_v = 2257 kJ/kg.The specific heat for ice is 2050 J/kg · K. )

A sealed 29 m³ tank is filled with 9000 moles of ideal oxygen gas (diatomic) at an initial temperature of 270 K.The gas is heated to a final temperature of 330 K.The atomic mass of oxygen is 16.0 g/mol.The initial pressure of the gas,in MPa,is closest to: