What is the final temperature and physical state of water when 250 grams of water at 85 degrees C is added to 80.0 grams of ice at -15 degree C?
Is the physical state of water liquid?
The explanation from bobpursly was this
The sum of heats gained =0
80(cice)(0+15)+80Hf+80cw(Tf-0)+250cw(Tf-85)=0
The 80 x specific heat ice x 15 = heat absorbed in changing the T of ice at -15 to zero C.
Then you melt the ice and convert solid ice to liquid water with 80 x heat fusion ice.
Then you combine the moving T to zero with melting and see if the 250 g water at 85 is enough to give you a positive temperature. I quickly looked at the figures and I think so. I appears Tfinal is about 55 C or so.
You can look at it another way.
heat gained by ice in moving from -15 to zero C is the first part.
Second part is melting the ice at zero to liquid water at zero.
Add those together.
Then calculate how much heat it is possible to lose with the 150 g H2O going from 85 to zero C. If you have more heat to lose than you gain by ice moving from -15 to zero then melting, you must have enough heat in the water to do that. If the heat in the 250 g H2O isn't enough then there isn't enough to melt all of the ice. Hope this helps.
To determine the final temperature and physical state of water when 250 grams of water at 85 degrees C is added to 80.0 grams of ice at -15 degrees C, we can use the principle of energy conservation.
First, let's define the unknowns:
- Tf = Final temperature of the system (water and ice) in degrees Celsius
- cice = Specific heat capacity of ice (0 degrees Celsius) in J/g°C
- Hf = Heat of fusion of ice in J/g
- cw = Specific heat capacity of water in J/g°C
The principle of energy conservation states that the sum of the heats gained and lost by each component in the system is equal to zero.
For the ice component:
Total heat gained = (mass of ice) x (specific heat of ice) x (final temperature - initial temperature)
Total heat gained = 80g x cice x (Tf - (-15))
For the phase change of ice:
Total heat gained = (mass of ice) x (heat of fusion of ice)
Total heat gained = 80g x Hf
For the water component:
Total heat gained = (mass of water) x (specific heat of water) x (final temperature - initial temperature)
Total heat gained = 250g x cw x (Tf - 85)
Applying the principle of energy conservation:
Total heat gained + Total heat gained + Total heat gained = 0
80g x cice x (Tf - (-15)) + 80g x Hf + 250g x cw x (Tf - 85) = 0
This equation allows us to solve for the final temperature, Tf. Once we have the value of Tf, we can determine the physical state of water by comparing it to the boiling point of water (100 degrees Celsius) and the freezing point of water (0 degrees Celsius). If Tf is greater than 100 degrees Celsius, the water is in a gaseous state (steam). If Tf is between 0 and 100 degrees Celsius, the water is in a liquid state. If Tf is less than 0 degrees Celsius, the water is in a solid state (ice).
So, to answer your question, we need to solve the equation given by bobpursly to find the final temperature (Tf), and then compare it to the boiling and freezing points of water to determine the physical state.