Thermodynamics B/C

[UTF-8 U+6211 U+662F]

When is the change in enthalpy equal to heat into the system?

Answer (Click to Show Hidden Text)

Constant pressure

Yep, your turn!

[Justin72835]

Yep, your turn!

An asteroid with a diameter of 10 km and a mass of 4.50 x 10^15 kg impacts the earth at a speed of 33.0 km/s. If 0.50% of the asteroid's kinetic energy goes into boiling the ocean water (assuming an initial water temperature of 10.0 degrees Celsius), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about 2 x 10^15 kg).

[UTF-8 U+6211 U+662F]

An asteroid with a diameter of 10 km and a mass of 4.50 x 10^15 kg impacts the earth at a speed of 33.0 km/s. If 0.50% of the asteroid's kinetic energy goes into boiling the ocean water (assuming an initial water temperature of 10.0 degrees Celsius), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about 2 x 10^15 kg).

Click to Show Answer

[Justin72835]

An asteroid with a diameter of 10 km and a mass of 4.50 x 10^15 kg impacts the earth at a speed of 33.0 km/s. If 0.50% of the asteroid's kinetic energy goes into boiling the ocean water (assuming an initial water temperature of 10.0 degrees Celsius), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about 2 x 10^15 kg).

Click to Show Answer

Correct! You're next!

[UTF-8 U+6211 U+662F]

What is the total entropy of a Carnot cycle between 50 K and 20 K?

[Justin72835]

What is the total entropy of a Carnot cycle between 50 K and 20 K?

Answer (Click to Show Hidden Text)

Isn't it just 0 J/K since it's a reversible cycle?

[UTF-8 U+6211 U+662F]

What is the total entropy of a Carnot cycle between 50 K and 20 K?

Answer (Click to Show Hidden Text)

Isn't it just 0 J/K since it's a reversible cycle?

Yes, your turn

[Justin72835]

Yes, your turn

Again, sorry for the late response. You take 780 grams of ice at a temperature of -12 degrees Celsius and mix it with 220 grams of steam at a temperature of 108 degrees Celsius. What is the common temperature once the two substances have met equilibrium?

[UTF-8 U+6211 U+662F]

Again, sorry for the late response. You take 780 grams of ice at a temperature of -12 degrees Celsius and mix it with 220 grams of steam at a temperature of 108 degrees Celsius. What is the common temperature once the two substances have met equilibrium?

Click to Show Answer

which after some fun algebra becomes

[Justin72835]

Again, sorry for the late response. You take 780 grams of ice at a temperature of -12 degrees Celsius and mix it with 220 grams of steam at a temperature of 108 degrees Celsius. What is the common temperature once the two substances have met equilibrium?

Click to Show Answer

which after some fun algebra becomes

Slight error (Click to Show Hidden Text)

Your work is essentially correct but you made a slight error by multiplying by the wrong mass:

Instead of:
[math]...+ 4187 \frac{J}{kg * K} * 0.78 kg * (100 \degree C - T_f)[/math]

It should be:
[math]...+ 4187 \frac{J}{kg * K} * 0.22 kg * (100 \degree C - T_f)[/math]

If you plug in the values with 0.22 kg instead of 0.78 kg, you get the correct answer of [b]73.11 degrees Celsius[/b]. However, you obviously knew what you were doing and it was probably just a mindless mistake  :shock: . You're next!