3. The quantity 0.850kg of ice at -20 degrees Celsius is added to 1.100kg of water in an aluminum Calorimetry cup at 7 degrees Celsius. The mass of the Calorimetry cup is 0.250kg. Use your knowledge of specific heats and latent heats to determine the temperature for the ice/water/cup system after it has come to thermal equilibrium. If that thermal equilibrium is zero degrees Celsius, determine how much of the ice in the cup has melted, and how much remains frozen. Another scenario for the zero degrees Celsius equilibrium is that where none of the ice melts, and instead some of the cooled water actually freezes, adding to the amount of ice in the system.
Note: This problem requires a careful analysis of both positive and negative quantities of heat.
Nuts to the note.
The sum of heats gained is zero. (some of the heats gained will be negative).
Heat gained by ice to zeroC+heatgained by melted ice at 0C +heat gained by water in cup+ heatgained by aluminum=0
Now the algebra:
.850*Cice*(0 - (-20)) + M*Hf+1.1cwater*(0-7)+.250Calum*(0-7)=0
where M is the mass of ice melting. Now if it comes out negative, that means ice was made from the water.
Hf is the latent heat of fusion of water/ice. It should be a heat gained.
My gut feel and head thinking...ignore the aluminum...is that ice is formed from the water, M will be negative, but it is close in my head. Happy computing
So am i supposed to know the latent heat of fusion of water and ice or is that something i look up/figure out? sorry physics is not my best subject...
To solve this problem, we need to consider the heat gains and losses in the system.
1. First, let's calculate the heat gained or lost by each component:
- Heat gained by the ice = mass of ice * specific heat of ice * change in temperature
Q_ice = 0.850kg * 2.09 J/g°C * (0 - (-20)°C)
- Heat gained by the water = mass of water * specific heat of water * change in temperature
Q_water = 1.100kg * 4.18 J/g°C * (0 - 7)°C
- Heat lost by the Calorimetry cup = mass of cup * specific heat of aluminum * change in temperature
Q_cup = 0.250kg * 0.89 J/g°C * (0 - 7)°C
2. Next, let's consider the heat gained/lost during the phase changes:
- Heat gained by the ice during melting = mass of melted ice * latent heat of fusion of water
Q_melt = mass of melted ice * 334 J/g
- Heat lost by the water during freezing = mass of frozen water * latent heat of fusion of water
Q_freeze = mass of frozen water * 334 J/g
3. Now, let's set up the equation for heat conservation:
Q_ice + Q_water + Q_cup + Q_melt - Q_freeze = 0
Substitute the above values into the equation and solve for the unknowns:
=> 0.850kg * 2.09 J/g°C * (0 - (-20)°C) + 1.100kg * 4.18 J/g°C * (0 - 7)°C + 0.250kg * 0.89 J/g°C * (0 - 7)°C + mass of melted ice * 334 J/g - mass of frozen water * 334 J/g = 0
4. Scenario 1: Zero degrees Celsius equilibrium with ice melting:
If the system reaches equilibrium at 0 degrees Celsius and some ice melts:
- Solve the equation and find the mass of melted ice.
- The remaining mass of ice in the cup will be the initial mass of ice minus the mass of melted ice.
5. Scenario 2: Zero degrees Celsius equilibrium with water freezing:
If the system reaches equilibrium at 0 degrees Celsius and none of the ice melts:
- Solve the equation and find the mass of frozen water.
- The total mass of ice in the cup will be the initial mass of ice plus the mass of frozen water.
By following these steps, you should be able to determine the temperature for the ice/water/cup system after thermal equilibrium and calculate the mass of melted ice or frozen water.