what is the final temperature of amixture containning 25 grams of ice at -20 degree celciusadded to water at room temperature 22oC mass of water is 50g
followed the steps given by you
but not getting the answer (0degree celsius)
ice at -20 to ice at 0 degree=mcdelta T=25*2.06*20=1030
1ce at 0 degree-water at 0=334*25 =8350
waterat 0 to water at final temp=25*4.18*tf-0
heat lost by water at 22 degree celsius=
50*4.18*(Tf-22)
1030+8350+104.5Tf=-(209Tf-4598)( negative since heat lost)
4782=313Tf
tf=15.3
http://www.jiskha.com/display.cgi?id=1399943202
All of your work is ok until here.
50*4.18*(Tf-22)
1030+8350+104.5Tf=-(209Tf-4598)( negative since heat lost)
4782=313Tf
tf=15.3
Should have been
1030 + 8350 + 104.5Tf + 209Tf + 4598 = 0
Then you get xxxTf = -some number and that makes Tf = -something and you KNOW that can't be right. What it means is that all of the ice didn't melt. If I didn't goof that other response should take care of it.
i don't understand
can you make it clear
If you followed your math you ended up with Tf of 15.3. I didn't go through line by line but my explanation was that your work should have produced a negative value for Tf. When that happens we know it can't be right from the data given so we know something is wrong. The wrong part is that I assumed all of the ice would melt and I gave the formula to you for that. Since all of the ice will not melt, that formula I gave you is no good. However, I worked the problem completely. If you can tell me what isn't clear perhaps I can help. But let's not beat a dead horse here. If it is something about how I worked the problem after knowing not all of the ice will melt perhaps I can help. If it's something about the first set up I used I think that is a waste of time since we already know that isn't the way to go.
To find the final temperature of the mixture, you need to calculate the amount of heat gained or lost by each component and add them up. Let's go through the steps again:
Step 1: Ice at -20°C to ice at 0°C
To calculate the heat gained by the ice, you use the formula: q = m * c * ΔT, where q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the mass of the ice is 25g, the specific heat capacity of ice is 2.06 J/g°C, and the change in temperature is 20°C. Therefore, the heat gained by the ice is:
q = 25g * 2.06 J/g°C * 20°C = 1030 J
Step 2: Ice at 0°C to Water at 0°C
To melt the ice at 0°C, it requires the heat of fusion. The formula for calculating the heat gained or lost during the phase change is: q = m * ΔH_fusion.
The heat of fusion for water is 334 J/g. Since the mass of the ice is still 25g, the heat gained during the phase change is:
q = 25g * 334 J/g = 8350 J
Step 3: Water at 0°C to the final temperature
To raise the temperature of the water at 0°C to the final temperature, we use the formula: q = m * c * ΔT.
The mass of the water is 50g, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is Tf - 0°C. Therefore, the heat gained by the water is:
q = 50g * 4.18 J/g°C * (Tf - 0°C) = 209Tf J (assuming Tf is the final temperature in °C)
According to the law of conservation of energy, the total heat gained by the system is equal to the total heat lost. So, by setting up an equation:
1030 J + 8350 J + 209Tf J = 50g * 4.18 J/g°C * (Tf - 22°C)
By simplifying the equation, we get:
4782 J + 209Tf J = 313Tf J - 6866 J
Combine like terms:
313Tf J - 209Tf J = 4782 J + 6866 J
104Tf J = 11648 J
Solve for Tf:
Tf = 11648 J / 104 J ≈ 112°C
It seems there may have been an error in your calculations or transcription, as the final temperature should be positive and not around 15.3°C. Please double-check the calculations to ensure accuracy.