Question 1

What is the net power radiated by a little animal with a surface area of 0.075 m² if his emissivity is 0.75, his skin temperature is 315 K, and he is in a room with a temperature of 290 K? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The net power radiated by the animal can be calculated using the Stefan-Boltzmann law, which is given by $P = e\sigma A(T^4 - T_0^4)$, where $e$ is the emissivity, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, $T$ is the temperature of the animal, and $T_0$ is the temperature of the surroundings. Plugging in the given values: $P = 0.75 \times 5.67 \times 10^{-8} \times 0.075 \times (315^4 - 290^4)$, we find that the correct answer is closest to 8.8 W, making option A the correct choice.

Question 2

The radius of a star is 6.95 × 10⁸ m, and its rate of radiation has been measured to be 5.32 × 10²⁶ W. Assuming that it is a perfect emitter, what is the temperature of the surface of this star? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The temperature of the star can be calculated using the Stefan-Boltzmann law: $P = 4\pi R^2 \sigma T^4$, where $P$ is the power (rate of radiation), $R$ is the radius of the star, $\sigma$ is the Stefan-Boltzmann constant, and $T$ is the temperature of the star. Rearranging for $T$, we get $T = \left(\frac{P}{4\pi R^2 \sigma}\right)^{1/4}$. Substituting the given values, $P = 5.32 \times 10^{26} \, \text{W}$, $R = 6.95 \times 10^{8} \, \text{m}$, and $\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4$, we find $T$ to be approximately $6.27 \times 10^{3} \, \text{K}$.

Question 3

Two metal rods, one silver and the other gold, are attached to each other end-to-end. The free end of the silver rod is immersed in a steam chamber at 100°C, and the free end of the gold rod in an ice water bath at 0°C. The rods are both 5.0 cm long and have a square cross-section that is 2.0 cm on a side. No heat is exchanged between the rods and their surroundings, except at the ends.
- What is the temperature at the point where the two rods are in contact with one another? The thermal conductivity of silver is 417 W/m ? K, and that of gold is 291 W/m ? K.

Accepted Answer

The answer of Two metal rods, one silver and the...

Question 4

Two identical objects are placed in a room with a temperature of 20°C. Object A has a temperature of 50°C, while object B has a temperature of 90°C. What is the ratio of the net power emitted by object B to the power emitted by object A?

Accepted Answer

The net power emitted by an object due to thermal radiation can be calculated using the Stefan-Boltzmann law, which states that the power emitted is proportional to the fourth power of the absolute temperature (in Kelvin). The formula for power emitted by an object is $P = \sigma A (T^4 - T_0^4)$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area of the object, $T$ is the temperature of the object, and $T_0$ is the temperature of the surroundings. Since both objects are identical and placed in the same environment, $A$, $\sigma$, and $T_0$ are constants for both objects. Therefore, the ratio of the net power emitted by object B to that of object A is given by the ratio of their temperatures to the fourth power, subtracting the fourth power of the room temperature in each case.Let's convert the temperatures to Kelvin by adding 273 to each:- Object A: $50°C = 323K$- Object B: $90°C = 363K$- Room temperature: $20°C = 293K$The ratio of the net power emitted is then:$$\frac{P_B}{P_A} = \frac{(363K)^4 - (293K)^4}{(323K)^4 - (293K)^4}$$Calculating this ratio gives a value close to 2.8, making option B the correct answer.

Question 5

A giant star radiates energy at the rate of 3.0 × 10³⁰ W, and its surface temperature has been measured to be 3000 K. Assuming that it is a perfect emitter, what is the radius of this star?(σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

We can use the Stefan-Boltzmann law to relate the power radiated by the star to its surface area and temperature:

P = σA T^4

where P is the power radiated, A is the surface area, T is the temperature, and σ is the Stefan-Boltzmann constant.

Solving for the surface area:

A = P/(σT^4)

Plugging in the given values:

A = (3.0 × 10^31 W)/(5.67 × 10^-8 W/m^2 K^4 * (3000 K)^4) = 2.32 x 10^19 m^2

Finally, we can use the formula for the surface area of a sphere to find its radius:

A = 4πr^2

r = sqrt(A/(4π)) = sqrt((2.32 x 10^19 m^2)/(4π)) ≈ 2.3 × 10^11 m

Therefore, the correct answer is (E) 2.3 × 10^11 m.

Question 6

How many molecules are in (a)1.0 cm³ of air at STP and (b)1.0 cm³ of helium at STP? (R = 8.31 J/mol . K, N_A = 6.022 × 10²³ molecules/mol)

Accepted Answer

The answer of How many molecules are in (a)1.0 cm³...

Question 7

How much power does a sphere with a radius of 10 cm radiate into empty space if is has an emissivity of 1.0 and is kept at a temperature of 400 K? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The power radiated by a black body (or a body with an emissivity of 1.0) is given by the Stefan-Boltzmann law:

P = σAT^4

where P is the power radiated, A is the surface area of the sphere (4πr^2), σ is the Stefan-Boltzmann constant, and T is the temperature in Kelvin.

Substituting the given values, we get:

P = (5.67 × 10^-8 W/m^2K^4) × (4π(0.1 m)^2) × (400 K)^4 ≈ 180 W

Therefore, the correct answer is C) 180 W.

Question 8

A jar holds 2.0 L of ideal nitrogen gas, N₂, at STP. The atomic mass of nitrogen is 14.0 g/mol, the ideal gas constant is R = 8.31 J/mol ? K, Avogadro's number is N_A = 6.022 × 10²³ molecules/mol, and 1.00 atm = 101 kPa. (a)How many moles of nitrogen are in the jar? (b)How many nitrogen molecules are in the jar? (c)What is the mass of the nitrogen in the jar?

Accepted Answer

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Question 9

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 with dimensions 8.00 cm × 8.00 cm. After 5.00 min of running the experiment, 64.0 g of ice have melted. What is the thermal conductivity of this material? The latent heat of fusion of water is 33.5 × 10⁴ J/kg, the latent heat of vaporization of water is 2.256 × 10⁶ J/kg, and both the ice and water are under 1.00 atm of pressure.

Accepted Answer

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Question 10

A sphere of surface area 1.25 m² and emissivity 1.0 is at a temperature of 100°C. At what rate does it radiate heat into empty space? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The answer of A sphere of surface area 1.25 m²...

Question 11

If a certain sample of an ideal gas has a temperature of 104°C and exerts a pressure of 2.3 × 10⁴ Pa on the walls of its container, how many gas molecules are present in each cubic centimeter of volume? The ideal gas constant is R = 8.31 J/mol × K and Avogadro's number is N_A = 6.022 × 10²³ molecules/mol.

Accepted Answer

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Question 12

The thermal conductivity of a certain concrete is 0.80 W/m . K and the thermal conductivity of a certain wood is 0.10 W/m ∙ K. How thick would a solid concrete wall have to be in order to have the same rate of heat flow through it as an 8.0-cm thick wall made of solid wood? Both walls have the same surface area and the same temperature difference across their faces.

Accepted Answer

The rate of heat flow is given by the formula Q/t = kA(T1-T2)/d, where Q is the heat flow rate, t is time, k is the thermal conductivity, A is the surface area, T1-T2 is the temperature difference, and d is the thickness of the wall. In this case, we want to find d for the concrete wall that will give the same heat flow rate as the 8.0-cm thick wood wall. Let's call this thickness d2. We can set up the equation:

0.80Ad1(T1-T2)/d1 = 0.10Ad2(T1-T2)/d2

Simplifying and solving for d2, we get:

d2 = 8.0*(0.80/0.10) = 64 cm

Therefore, the answer is B, 64 cm.

Question 13

A rod, with sides insulated to prevent heat loss, has one end immersed in boiling water at 100°C and the other end in a water-ice mixture at 0°C. The rod has uniform cross-sectional area $7.05 \mathrm {~cm} ^ { 2 }$ and length $87 \mathrm {~cm} .$ The heat conducted by the rod melts the ice at a rate of 1.0 g every 11 seconds. What is the thermal conductivity of the rod? Recall that the heat of fusion of water is 3.34 × 10⁵ J/kg.

Accepted Answer

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Question 14

The cylindrical filament in a light bulb has a diameter of 0.050 mm, an emissivity of 1.0, and a temperature of 3000°C. How long should the filament be in order to radiate 60 W of power? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The answer of The cylindrical filament in a light bulb...

Question 15

The thermal conductivity of aluminum is twice that of brass. Two rods (one aluminum and the other brass)of the same length and cross-sectional area are joined together end to end. The free end of the brass rod is maintained at 0°C and the free end of the aluminum rod is maintained at 200°C. If no heat escapes from the sides of the rods, what is the temperature at the place where the two rods are joined together?

Accepted Answer

Let T be the temperature at the point where the two rods are joined. The rate of heat flow through both rods must be the same.
Thus, (T - 0)/L = 2(200 - T)/L
Simplifying this equation, we get:
T = 133°C
Therefore, the answer is choice B.

Question 16

What is the net power that a person with surface area of 1.20 m² radiates if his emissivity is 0.895, his skin temperature is 27°C, and he is in a room that is at a temperature of 17°C? (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

We can use the formula for net power radiated by a person:
P = εσA(T_s^4 - T_r^4)
where
ε = emissivity = 0.895
σ = Stefan-Boltzmann constant = 5.67 × 10^-8 W/m^2K^4
A = surface area = 1.20 m^2
T_s = skin temperature = 27 + 273 = 300 K
T_r = room temperature = 17 + 273 = 290 K
Plugging in the numbers, we get:
P = 0.895 × 5.67 × 10^-8 × 1.20 × (300^4 - 290^4) = 62.6 W
Therefore, the answer is B.

Question 17

Originally 2.00 mol of gas are at STP. If the temperature changes to 47.0°C and the pressure decreases to half of what it was, how many liters do the two moles now occupy? (1 atm = 101 kPa, R = 8.31 J/mol ? K)

Accepted Answer

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Question 18

A blacksmith is flattening a steel plate having dimensions 10 cm × 15 cm × 1 mm. He has heated the plate to 900 K. If the emissivity of the plate is 0.75, at what rate does it lose energy by radiation? Ignore any heat exchange with the surroundings. (σ = 5.67 × 10^-8 W/m² ∙ K⁴)

Accepted Answer

The answer of A blacksmith is flattening a steel plate...

Question 19

A window glass that is 0.5 cm thick has dimensions of 3 m by 1.5 m. The thermal conductivity of this glass is 0.8 W/m ∙ K. If the outside surface of the glass is at -10°C and the inside surface is at 20°C, how much heat flows through the window in every hour?

Accepted Answer

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Question 20

In an electric furnace used for refining steel, the temperature is monitored by measuring the radiant power emitted through a small hole in the wall of the furnace, of area 0.5 cm². This hole acts like a perfect blackbody radiator having the same temperature as the interior of the furnace. If the temperature of the furnace (and therefore of the hole)is to be maintained at 1650°C, how much power will the hole radiate?

Accepted Answer

The power radiated from a black body is given by the Stefan-Boltzmann law:
P = σAT^4
where P is the radiant power, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the area of the hole (0.5 cm^2 = 5 x 10^-6 m^2), and T is the temperature in Kelvin (1650 + 273 = 1923 K).

Substituting the values, we get:
P = (5.67 x 10^-8)(5 x 10^-6)(1923^4)
P ≈ 40 W

Therefore, the power radiated from the hole is approximately 40 W. So, the correct choice is C) 40 W.

Quiz 12: Thermal Properties of Matter

What is the net power radiated by a little animal with a surface area of 0.075 m2 if his emissivity is 0.75, his skin temperature is 315 K, and he is in a room with a temperature of 290 K? (σ = 5.67 × 10-8 W/m2 ∙ K4)

The radius of a star is 6.95 × 108 m, and its rate of radiation has been measured to be 5.32 × 1026 W. Assuming that it is a perfect emitter, what is the temperature of the surface of this star? (σ = 5.67 × 10-8 W/m2 ∙ K4)

Two identical objects are placed in a room with a temperature of 20°C. Object A has a temperature of 50°C, while object B has a temperature of 90°C. What is the ratio of the net power emitted by object B to the power emitted by object A?

A giant star radiates energy at the rate of 3.0 × 1030 W, and its surface temperature has been measured to be 3000 K. Assuming that it is a perfect emitter, what is the radius of this star?(σ = 5.67 × 10-8 W/m2 ∙ K4)

How many molecules are in (a)1.0 cm3 of air at STP and (b)1.0 cm3 of helium at STP? (R = 8.31 J/mol . K, NA = 6.022 × 1023 molecules/mol)

How much power does a sphere with a radius of 10 cm radiate into empty space if is has an emissivity of 1.0 and is kept at a temperature of 400 K? (σ = 5.67 × 10-8 W/m2 ∙ K4)

A sphere of surface area 1.25 m2 and emissivity 1.0 is at a temperature of 100°C. At what rate does it radiate heat into empty space? (σ = 5.67 × 10-8 W/m2 ∙ K4)

The cylindrical filament in a light bulb has a diameter of 0.050 mm, an emissivity of 1.0, and a temperature of 3000°C. How long should the filament be in order to radiate 60 W of power? (σ = 5.67 × 10-8 W/m2 ∙ K4)

What is the net power that a person with surface area of 1.20 m2 radiates if his emissivity is 0.895, his skin temperature is 27°C, and he is in a room that is at a temperature of 17°C? (σ = 5.67 × 10-8 W/m2 ∙ K4)