A pressure gauge indicates the differences between atmospheric pressure and pressure inside the tank. The gauge on a 2.00 m^3 oxygen tank reads 36 atm. After some use of the oxygen, the gauge reads 24 atm. How many cubic meters of oxygen at normal atmospheric pressure was used? There is no temperature change during the time of consumption.

Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm.
Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3.
You can go ahead with the next question though :D

Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.

Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm.
Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3.
You can go ahead with the next question though :D

Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.

Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm.
Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3.
You can go ahead with the next question though :D

Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.

Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.

Yep! Although you could explain the last part a little more (non sequitur there). Your turn!

Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.

Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.

Yep! Although you could explain the last part a little more (non sequitur there). Your turn!

Alright. A cup of tea cools from 170 degrees Fahrenheit to 145 degrees Fahrenheit in 6 minutes in a room at 72 degrees Fahrenheit. How long will it take for the tea to cool from 100 to 90 degrees Fahrenheit in the same room?

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

Alright. A cup of tea cools from 170 degrees Fahrenheit to 145 degrees Fahrenheit in 6 minutes in a room at 72 degrees Fahrenheit. How long will it take for the tea to cool from 100 to 90 degrees Fahrenheit in the same room?

Using Newton's Law of Cooling, we get T = Ts + (T0-Ts)e^kt. Plugging in our data gives us a k value of -4.91E-2. We can set up this equation again with 100 to 90 degrees, and calculate a t of 9.0 minutes.

Alright. A cup of tea cools from 170 degrees Fahrenheit to 145 degrees Fahrenheit in 6 minutes in a room at 72 degrees Fahrenheit. How long will it take for the tea to cool from 100 to 90 degrees Fahrenheit in the same room?

Using Newton's Law of Cooling, we get T = Ts + (T0-Ts)e^kt. Plugging in our data gives us a k value of -4.91E-2. We can set up this equation again with 100 to 90 degrees, and calculate a t of 9.0 minutes.

Great job! Your turn!

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

A copper can, of negligible heat capacity, contains 1.000 kg of water just above the freezing point. A similar can contains 1.000 kg of water just below the boiling point. The two cans are brought into thermal contact. Find the change in entropy of the system. (Assume for water cm = 4184 J/K)

A copper can, of negligible heat capacity, contains 1.000 kg of water just above the freezing point. A similar can contains 1.000 kg of water just below the boiling point. The two cans are brought into thermal contact. Find the change in entropy of the system. (Assume for water cm = 4184 J/K)

Too lazy to put work, but since the masses are the same, you can use the equation mcln(Tf/Ti) for both volumes of water to get an answer of 101J/K.

"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

University of Texas at Austin '23
Seven Lakes High School '19

A copper can, of negligible heat capacity, contains 1.000 kg of water just above the freezing point. A similar can contains 1.000 kg of water just below the boiling point. The two cans are brought into thermal contact. Find the change in entropy of the system. (Assume for water cm = 4184 J/K)

Too lazy to put work, but since the masses are the same, you can use the equation mcln(Tf/Ti) for both volumes of water to get an answer of 101J/K.