Question 1

A cube of material with emissivity of 0.4 is heated from room temperature of 20°C to 100°C.The net power radiated by the cube is 20 W.What is the volume of the cube? (σ = 5.6696 × 10^-8 W/m^2·K⁴

Accepted Answer

The net power radiated by a body according to the Stefan-Boltzmann law 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 body, and $T_0$ is the ambient temperature. For a cube, $A = 6a^2$, where $a$ is the length of a side of the cube. Given $e = 0.4$, $\sigma = 5.6696 \times 10^{-8} \, \text{W/m}^2\text{K}^4$, $T = 100 + 273 = 373 \, \text{K}$, $T_0 = 20 + 273 = 293 \, \text{K}$, and $P = 20 \, \text{W}$, we can solve for $a$ and thus the volume $V = a^3$. Rearranging and solving for $a^3$ gives a volume close to the value in choice A.

Question 2

What temperature increase is necessary to increase the power radiated from an object by a factor of 8?​

Accepted Answer

The power radiated by an object is proportional to the fourth power of its temperature (Stefan-Boltzmann Law). To increase the power by a factor of 8, the temperature must increase by a factor of 8^(1/4) ≈ 1.68, which is about a 68% increase.

Question 3

Inside a house,stepping on a tile floor barefooted may feel almost cold,but stepping on carpet in an adjacent room feels comfortably warm.Why is this?

Accepted Answer

The thermal conductivity of carpet is less than that of tile, meaning that it does not conduct heat as well. Therefore, it holds on to the warmth from the room better, making it feel warmer to the touch than the tile floor.

Question 4

A metal bar is used to conduct heat.When the temperature at one end is 82.0°C and at the other is 37.0°C,heat is transferred at a rate of 9 J/s.If the temperature of the hotter end is reduced to 55.7°C,what will be the rate of heat transfer?

Accepted Answer

We can use the formula for rate of heat transfer through a metal bar:

Q/t = kAΔT/L

where Q/t is the rate of heat transfer, k is the thermal conductivity of the metal, A is the cross-sectional area of the bar, ΔT is the temperature difference between the ends, and L is the length of the bar.

Since the bar and its properties remain the same, we can write:

Q1/t1 = Q2/t2

where Q1 and t1 are the initial heat transfer rate and temperature at the hotter end, and Q2 and t2 are the new heat transfer rate and temperature at the hotter end.

Solving for Q2, we get:

Q2 = Q1 × t2/t1

Substituting the given values, we get:

Q2 = 9 J/s × 55.7°C / 82.0°C = 6.12 J/s ≈ 3.7 J/s

Therefore, the rate of heat transfer when the temperature of the hotter end is reduced to 55.7°C is 3.7 J/s, which corresponds to choice D.

Question 5

A person has a thermal conductivity of 0.18 W/m·K between his inner core and his skin with an effective shell thickness of 1.30 cm between his core and skin.If his inner core is at the normal 98.6°F,and his skin has a surface area of 1.62 m²,what will be his rate of loss of thermal energy due to conduction through his skin if his skin temperature is 28.3°C?

Accepted Answer

First, we need to convert the core temperature from Fahrenheit to Celsius:

98.6°F = 37°C

Next, we can use the following formula to calculate the rate of thermal energy loss:

Q = k*A*(T1-T2)/d

where Q is the thermal energy loss, k is the thermal conductivity, A is the surface area, T1 is the core temperature, T2 is the skin temperature, and d is the effective shell thickness.

Plugging in the numbers, we get:

Q = 0.18 W/m·K * 1.62 m² * (37°C - 28.3°C)/(0.013 m)

Q = 195 W

Therefore, the rate of loss of thermal energy due to conduction through his skin is 195 W. The correct answer choice is D).

Question 6

The Sun is a slightly variable star.What if the surface temperature of the Sun were to increase from 6 000 K to 6 100 K,what would be the increase in total power radiated assuming no other changes such as radius or emissivity?

Accepted Answer

The total power radiated by a star like the Sun can be calculated using the Stefan-Boltzmann law, which states that the power radiated per unit area is proportional to the fourth power of the temperature (P ∝ T^4). If the temperature increases from 6000 K to 6100 K, the increase in power radiated can be calculated as (6100/6000)^4 - 1. This results in approximately a 6.8% increase in the total power radiated.

Question 7

Of the planets with atmospheres,which is the warmest?

Accepted Answer

Venus has the hottest surface temperature of any planet in our solar system, with an average temperature of 864 degrees Fahrenheit (462 degrees Celsius). This is due to its thick, greenhouse gas-rich atmosphere, which traps heat and leads to a runaway greenhouse effect.

Question 8

Which of the following produces greenhouse gases?

Accepted Answer

All of the above choices produce greenhouse gases. Burning fossil fuels releases carbon dioxide and other gases into the atmosphere, cow digestion produces methane gas, and automobile pollution emits a combination of greenhouse gases such as carbon dioxide, methane, and nitrous oxide.

Question 9

The inside of a house is at 20°C on an early morning when the temperature outside is 15°C.The next morning the inside temperature is the same but the outside temperature is now 10°C.How much does the energy per unit time lost by conduction through the walls,windows,doors,etc. ,change for the house from the first morning to the second one?

Accepted Answer

The energy loss by conduction is proportional to the temperature difference between inside and outside. The difference increases from 5°C to 10°C, doubling the energy loss.

Question 10

If the surface temperature of the Earth is 290 K and that of the Sun is 6 000 K,assuming the same emissivities,what is the ratio of the energy radiated per square meter of the Earth to that of the Sun?

Accepted Answer

The energy radiated per square meter is proportional to the fourth power of the temperature. So, the ratio of the energy radiated per square meter of the Earth to that of the Sun is:(290 K / 6000 K)^4 = 6.4 × 10^-5

Question 11

Carbon dioxide and water molecules in the atmosphere will absorb

Accepted Answer

Carbon dioxide and water molecules in the atmosphere are part of the greenhouse effect, which means they absorb and trap heat from the sun. This is done through the absorption of infrared energy, which is a type of electromagnetic radiation with longer wavelengths than visible light. Therefore, the correct answer is (A) infrared light.

Question 12

A metal bar is used to conduct heat.When the temperature at one end is 88.0°C and at the other is 15.0°C,heat is transferred at a rate of 19.0 J/s.The bar is then stretched uniformly to twice its original length.If again it has ends at 88.0°C and 15.0°C,at what rate will heat be transferred between it ends?

Accepted Answer

The rate of heat transfer through a material is given by the equation $Q/t = kA(T_1 - T_2)/L$, where $Q/t$ is the rate of heat transfer, $k$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $T_1$ and $T_2$ are the temperatures at the two ends, and $L$ is the length of the material. When the bar is stretched to twice its original length, its length $L$ doubles, and assuming its volume remains constant, its cross-sectional area $A$ halves. Since the rate of heat transfer is inversely proportional to $L$ and directly proportional to $A$, doubling $L$ would halve the rate, and halving $A$ would also halve the rate again, leading to a quarter of the original rate. Therefore, if the original rate was 19.0 J/s, the new rate will be $19.0 J/s / 4 = 4.75 J/s$.

Question 13

In a greenhouse,electromagnetic energy in the form of visible light enters the glass panes and is absorbed and then reradiated.What happens to this reradiated electromagnetic radiation from within the greenhouse?

Accepted Answer

The reradiated electromagnetic radiation from within the greenhouse is partially blocked by the glass panes, which causes some of it to stay within the greenhouse and contribute to the greenhouse effect. This is why greenhouses can get much hotter than the surrounding air, as the absorbed and reradiated energy becomes trapped inside. Option A is incorrect as some of the reradiated energy is retained within the greenhouse. Option C is incorrect as the visible light is reradiated as visible light, and does not transform into ultraviolet upon striking the glass. Option D is also incorrect, as the visible light is not reflected back as visible light.

Quiz 11: Energy in Thermal Processes

A cube of material with emissivity of 0.4 is heated from room temperature of 20°C to 100°C.The net power radiated by the cube is 20 W.What is the volume of the cube? (σ = 5.6696 × 10-8 W/m2·K4