## Thermodynamics B/C

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### Re: Thermodynamics B/C

Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:When is the change in enthalpy equal to heat into the system?
Constant pressure

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote: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).
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### Re: Thermodynamics B/C

Justin72835 wrote: 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).
$KE = 0.5 * 4.50*10^{15} kg * (33.0 \frac{km}{s})^2$
$0.0050 * 0.5 * 4.50*10^{15} kg * (33.0 \frac{km}{s})^2 = m * 4.184 \frac{kJ}{kg * K} * 90.0 K + m * 2230 kJ/kg$
$1.2251*10^{16} \frac{kg*km^2}{s^2} = m(376.6 \frac{kJ}{kg} + 2230 \frac{kJ}{kg})$
$1.2251*10^{19} kJ = m(2606.6 \frac{kJ}{kg})$
$m = 4.7*10^{15} kg$

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote: 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).
$KE = 0.5 * 4.50*10^{15} kg * (33.0 \frac{km}{s})^2$
$0.0050 * 0.5 * 4.50*10^{15} kg * (33.0 \frac{km}{s})^2 = m * 4.184 \frac{kJ}{kg * K} * 90.0 K + m * 2230 kJ/kg$
$1.2251*10^{16} \frac{kg*km^2}{s^2} = m(376.6 \frac{kJ}{kg} + 2230 \frac{kJ}{kg})$
$1.2251*10^{19} kJ = m(2606.6 \frac{kJ}{kg})$
$m = 4.7*10^{15} kg$
Correct! You're next!
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### Re: Thermodynamics B/C

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

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:What is the total entropy of a Carnot cycle between 50 K and 20 K?
Isn't it just 0 J/K since it's a reversible cycle?
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### Re: Thermodynamics B/C

Justin72835 wrote:
UTF-8 U+6211 U+662F wrote:What is the total entropy of a Carnot cycle between 50 K and 20 K?
Isn't it just 0 J/K since it's a reversible cycle?

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote: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?
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### Re: Thermodynamics B/C

Justin72835 wrote: 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?
$Q_{ice} + Q_{fusion} + Q_{heatingwater} = Q_{steam} + Q_{vaporization} + Q_{coolingwater}$
$2108 \frac{J}{kg * K} * 0.78 kg * 12 K + 334000 \frac{J}{kg} * 0.78 kg + 4187 \frac{J}{kg * K} * 0.78 kg * (T_f - 0 \degree C) = 1996 \frac{J}{kg * K} * 0.22 kg * 8 K + 2230000 \frac{J}{kg} * 0.22 kg + 4187 \frac{J}{kg * K} * 0.78 kg * (100 \degree C - T_f)$
which after some fun algebra becomes
$T_f = 82.74 \degree C$

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote: 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?
$Q_{ice} + Q_{fusion} + Q_{heatingwater} = Q_{steam} + Q_{vaporization} + Q_{coolingwater}$
$2108 \frac{J}{kg * K} * 0.78 kg * 12 K + 334000 \frac{J}{kg} * 0.78 kg + 4187 \frac{J}{kg * K} * 0.78 kg * (T_f - 0 \degree C) = 1996 \frac{J}{kg * K} * 0.22 kg * 8 K + 2230000 \frac{J}{kg} * 0.22 kg + 4187 \frac{J}{kg * K} * 0.78 kg * (100 \degree C - T_f)$
which after some fun algebra becomes
$T_f = 82.74 \degree C$
Your work is essentially correct but you made a slight error by multiplying by the wrong mass:

$...+ 4187 \frac{J}{kg * K} * 0.78 kg * (100 \degree C - T_f)$

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

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

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