posted by Tichele on .
If 50.0 kJ of heat are added to a 20.0 gram ice cube at -5.00 degrees Celsius, what will be the resulting state and temperature of the water?
I need some serious help with this! I think I'm making it much harder than it really has to be.
OK. You have 50,000 Joules of heat.
How much heat do you need to move the temperature of the ice at -5.00 C to zero C?
q = mass ice x specific heat ice x (0-(-5)) = ??. Solve for q and subtract joules from 50,000. What's left?
Now how much does it take to melt 20 g ice?
That's q = mass x heat fusion.
Solve for q and see if you have that much heat left from the first part. If so all of the ice will melt. If not, some of it will melt and you can determine how much from the numbers.
Post your work if you get stuck and it would help if you included the heat fusion and the specific heat ice.
All right. I'll go try and work on this problem and come back if I need some help.
The specific heat of ice: 2.092 J/g-Kelvin
The specific heat of water: 4.184 J/g-Kelvin
The specific heat of steam: 1.841 J/g-Kelvin
Heat of Fusion: 6.008 kJ/mol at 0.00 degrees Celsius
Heat of Vaporization: 40.67 kJ/mol at 100 degrees Celsius
Ok, so I'm stuck....Here is my work thus far (please see my previous response for the values of fusion and vaporization, as well as the specific heats)
q=m x specific heat x change in temp
=20.0 g x (2.o92 J/g x Kelvin) x (0--5)=209.2 J
So... 50,000 J - 209.2 J= 49790.8 J left over
so now for the heat of fusion
q= mass x heat fusion= (20.0 g/18g/mol) x 6.006 kJ/mol= 6.675 kJ which is 6675 J
By now, the ice is melted and we still have 43115.8 J left
q=m x specific heat x change in temp for water
=20.0 g x (4.184 J/g x Kelvin) x 100 C=8368 J
This leaves me with 34747.8 J of heat
But if I try to work in the heat of evaporation, I get 45188.8 J and that would leave me with a negative number. I'm not sure where to go from here to find what Temperature my resulting state would be in...
So we start with 50,000 joules heat.
mass ice x specific heat ice x delta T = 20 g x 2.092 J/g x 5 = 209.2 which is much less than 50,000 and all of the ice will move from T = -5 to T = 0 C.
To change ice at zero C to liquid water at zero C we newed
mass ice x heat fusion
20g x 333.5 J/g (I changed 6.008 kJ/mol to J/g) = 6,670 J which means we have enough heat to melt all of the ice. How much heat is left?
50,000 - 209.2 - 6,670 = ??
So the extra J remaining will go to heat the water at zero from the melted ice to some higher T. How much heat is needed to heat the water at zero C to 100 C?
20g x 4.184 J/g*C x 100 = 8368 J.
How much is left now?
50,000 - 209.2-6670-8368 = 34,752.8 J.
You were showing your work while I was showing mine so I'll try to catch up. You're right, your last calculation leaves you with a negative number which means that not all of the water can be vaporized. The easy way to know how much CAN be vaporized is to use that same formula
mass water x 2257.56 J/g = 34752.8 ans solve for mass water. I get something like 15 or so.
So 15.394 g H2O x 2257.56 = which checks out.
That will leave 20.00-15.394 = ??g H2O at 100 C and 15.394 g steam at 100 C.
I don't think we are allowed that many significant figures but we can't make sure the heat works out without using all of them. You should round at the end.
I'm working on this same problem
I'm right there with you until the last step where you have Mass Water x 2257.56 = 34752.8
Can you explain to me where the 2257.56 comes from?