Question 1

During an isothermal process, 5.0 J of heat are removed from an ideal gas. What is the change in internal energy?

Accepted Answer

In an isothermal process, there is no change in the internal energy of an ideal gas. The internal energy is a function of temperature only, and since the temperature is constant, the internal energy remains unchanged. Therefore, the answer is zero.

Question 2

A heat pump uses 100 J of work to output 150 J of heat at some temperature. What is the coefficient of performance for the heat pump?

Accepted Answer

The coefficient of performance (COP) for a heat pump is defined as the ratio of the heat output to the work input. In this case, the COP is 150 J / 100 J = 1.5.

Question 3

A heat pump uses 100 J of work to output heat at some temperature. If the heat pump withdraws 60 J of heat from the lower temperature reservoir, then what is the coefficient of performance for the heat pump?

Accepted Answer

The answer of A heat pump uses 100 J of...

Question 4

A refrigerator uses 40.0 J of work to extract 90.0 J from a heat reservoir at 0.00°C. What is the coefficient of performance for the refrigerator?

Accepted Answer

The coefficient of performance (COP) for a refrigerator is given by the formula COP = Qc/W, where Qc is the heat extracted from the cold reservoir (90.0 J in this case) and W is the work input (40.0 J). Therefore, COP = 90.0 J / 40.0 J = 2.25.

Question 5

A heat pump uses 100 J of work to output 150 J of heat at some temperature. What is the heat that is withdrawn from some lower temperature?

Accepted Answer

The heat withdrawn from a lower temperature is the difference between the heat output and the work input, which is 150 J - 100 J = 50 J.

Question 6

A engine goes through a cyclic process. During the cyclic process the engine does 600 J of work and has a heat output of 2,400 J. What is the efficiency of the engine?

Accepted Answer

The efficiency of an engine is given by the formula efficiency = (work output / heat input) * 100%. In this case, the work output is 600 J and the heat input is 2,400 J. Therefore, the efficiency = (600 J / 2,400 J) * 100% = 25%. However, since 25% is not an option and there seems to be a mistake in my calculation, the closest correct option based on the provided formula and typical understanding of efficiency in thermodynamics would be 20%, assuming a possible error in the calculation or interpretation of the options.

Question 7

A heat engine has an efficiency of 25.0% and a power output of 600 W. What is the rate of heat input?

Accepted Answer

Efficiency = Output / Input*100
25% = 600 / Input*100
Input = 2400 W = 2.4 kW

Question 8

A engine goes through a cyclic process. During the cyclic process the engine takes in 3,200 J of heat and has a heat output of 2,400 J. What is the efficiency of the engine?

Accepted Answer

Efficiency = (Heat output / Heat input) * 100%
Efficiency = (2400 J / 3200 J) * 100% = 0.75 * 100% = 75%
Therefore, the efficiency of the engine is 25%. Option C is the closest answer to this value.

Question 9

A very efficient engine has the following characteristics-combustion = 1,900°C, exhaust = 430°C, 5.0 × 10⁶ cal of fuel produces 1.4 × 10⁷ J of work in one hour. What is the output in hp? (1 hp = 745.7 W)

Accepted Answer

The answer of A very efficient engine has the following...

Question 10

During an isothermal process, 5.0 J of heat are removed from an ideal gas. What is the work done by the gas in the process?

Accepted Answer

In an isothermal process, the temperature of the gas remains constant. Therefore, the internal energy of the gas also remains constant. According to the first law of thermodynamics, ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to or removed from the system, and W is the work done by or on the system. Since ΔU = 0 in an isothermal process, we can rearrange the equation to get W = Q. Thus, the work done by the gas is equal in magnitude but opposite in sign to the heat removed, which is -5.0 J. Therefore, the answer is B.

Question 11

An engine takes in 150 J of energy from a 400 K thermal source. If the engine discharges energy into a thermal reservoir at 150 K, then what is the maximum amount of work one can get out of the engine?

Accepted Answer

The maximum amount of work that can be extracted from an engine operating between two thermal reservoirs is given by the efficiency of an ideal Carnot engine, which is $1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$, where $T_{\text{cold}}$ and $T_{\text{hot}}$ are the temperatures of the cold and hot reservoirs, respectively. In this case, $T_{\text{hot}} = 400 \, K$ and $T_{\text{cold}} = 150 \, K$. The efficiency is $1 - \frac{150}{400} = 0.625$. The maximum work output is then the product of this efficiency and the energy input, which is $150 \, J \times 0.625 = 93.75 \, J$, which rounds to $93.8 \, J$.

Question 12

A heat engine has an efficiency of 35.0% and receives 150 J of heat per cycle. How much work does it perform in each cycle?

Accepted Answer

The work performed by the heat engine in each cycle can be calculated using the efficiency formula: Efficiency = Work Output / Heat Input. Given an efficiency of 35.0%, or 0.35, and a heat input of 150 J, the work output is 0.35 * 150 J = 52.5 J.

Question 13

In an isochoric process, the internal energy of a system decreases by 50 J. What is the work done on the system?

Accepted Answer

In an isochoric process, the volume of the system remains constant, so there is no work done by or on the system (W = 0). The change in internal energy (∆U) is solely due to heat transfer (Q) or other internal processes, such as chemical reactions or phase changes. Therefore, the work done on the system is zero (Choice A).

Question 14

An engine delivers 40.0 J of work while it takes in energy from a 400 K thermal source. If the engine discharges energy into a thermal reservoir at 200 K, then what is the least amount of heat needed to run the engine?

Accepted Answer

The efficiency of a heat engine operating between two thermal reservoirs is given by $1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$, where $T_{\text{cold}}$ and $T_{\text{hot}}$ are the temperatures of the cold and hot reservoirs, respectively. In this case, $T_{\text{cold}} = 200 \, \text{K}$ and $T_{\text{hot}} = 400 \, \text{K}$, so the maximum efficiency is $1 - \frac{200}{400} = 0.5$ or 50%. Since the engine delivers 40.0 J of work, the least amount of heat needed to run the engine, given its maximum efficiency, would be the work output divided by the efficiency: $40.0 \, \text{J} / 0.5 = 80.0 \, \text{J}$.

Question 15

When the first law of thermodynamics is applied to an ideal gas that is taken through an isothermal process

Accepted Answer

For an ideal gas undergoing an isothermal process, the change in internal energy (ΔU) is zero because internal energy is a function of temperature, which remains constant in an isothermal process.

Question 16

A heat engine has an efficiency of 35.0% and receives 150 J of heat per cycle. How much heat does it exhaust in each cycle?

Accepted Answer

The efficiency formula for a heat engine is: efficiency = (work output/heat input) x 100%
Since the work output is not given in the problem, we can use the fact that the work output is equal to the difference between the heat input and the heat exhausted. That is, work output = heat input - heat exhausted.
Substituting these values into the efficiency formula, we get:
35.0% = ((heat input - heat exhausted)/heat input) x 100%
0.35 = (heat input - heat exhausted)/heat input
Heat input = 150 J
0.35 x 150 J = heat input - heat exhausted
Heat exhausted = 52.5 J
Therefore, the answer is choice C, 97.5 J.

Question 17

In an isochoric process, the internal energy of a system decreases by 50 J. What is the heat flow to the system?

Accepted Answer

In an isochoric process (constant volume), no work is done by the system since the volume does not change. Therefore, any change in internal energy is due solely to heat flow. If the internal energy decreases by 50 J, it means 50 J of heat has left the system, hence the heat flow to the system is -50 J.

Question 18

A heat pump uses 200 J of work to output 300 J of heat into a heat reservoir at 200 K. What is the coefficient of performance for the heat pump?

Accepted Answer

The coefficient of performance (COP) for a heat pump is given by the formula COP = Q_h / W, where Q_h is the heat delivered to the hot reservoir and W is the work input. Here, Q_h = 300 J and W = 200 J, so COP = 300 J / 200 J = 1.5.

Question 19

10 joules of heat are transferred to a sample of ideal gas at constant volume. As a result, the internal energy of the gas

Accepted Answer

For an ideal gas at constant volume, all the heat transferred to the gas goes into increasing its internal energy, hence the internal energy increases by the same amount as the heat transferred, which is 10 J.

Question 20

A refrigerator uses 40 J of work to extract heat from a heat reservoir at 0.00°C. If the coefficient of performance of the refrigerator is 2.1, then how much heat is extracted from the heat reservoir at 0.00°C?

Accepted Answer

The coefficient of performance (COP) of the refrigerator is given by COP = heat removed from the refrigerated space / work done on the refrigerator. In this case, COP = 2.1, so the heat removed from the refrigerated space is 2.1 times the work done on the refrigerator.

Since the refrigerator is extracting heat from a reservoir at 0.00°C, we know that the heat removed is equal to the change in entropy of the reservoir divided by its temperature. Using the specific heat capacity of water, we can calculate the change in entropy for a given amount of heat.

ΔS = Q / T = (40 J) / (273 K) = 0.147 J/K

The heat removed by the refrigerator is then:

Q = COP x work done = 2.1 x 40 J = 84 J

Therefore, the best choice is C (84 J) as this is the amount of heat that is extracted from the heat reservoir at 0.00°C.

Quiz 15: Thermodynamics