If you mixed 200g of ice that is at -5°C with 20 g of water that is at 15°C, what will be the temperature and condition of the final state once equilibrium is achieved?
-
First assume the final temp (T) is above 0ºC. If not we have to recalculate. I suspect the latter.
Then the heat required to warm the ice to 0º and to melt the ice and then to warm the ice water to T
is equal to the heat required to cool the water from 15ºC to T.
specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
to warm ice to T
E1 = 2.06 kJ/kgC x 0.2 kg x 5C
to melt ice
E2 = 334 kJ/kg x 0.2 kg
to warm water from 0C to T
E3 = 4.186 kJ/kgC x 0.2 kg x (T–0)
Total energy taken from the ice is
E4 = 2.06 kJ + 66.8 kJ + 0.837T = 68.86 kJ + 0.837T
to cool water
E5 = 4.186 kJ/kgC x 0.02 kg x (15–T)
E5 = 0.0837(15–T) = 1.2558 – 0.0837T
set them equal and solve for T
68.86 + 0.837T = 1.2558 – 0.0837T
0.837T + 0.0837T = 1.2558 – 68.86
this is negative, so not all the ice melts.
Starting over, there are two new possibilities.
1. the ice warms up to the melting point and some of it melts.
2. The ice doesn't warm up to the melting point.
In case 1, final temp is 0º
case 2, T<0º
to warm ice to T (a negative number)
change in temperature is T+5
E1 = 2.06 kJ/kgC x 0.2 kg x (T+5)
E1 = 10.3T + 51.5
To cool water to 0º
E2 = 4.186 kJ/kgC x 0.02 kg x 15 = 1.256 kJ
to freeze the water
E3 = 334 kJ/kg x 0.02 kg = 6.68 kJ
to cool water from 0 to T
E4 = 2.06 kJ/kgC x 0.02 kg x (T+5)
E4 = 0.0412T + 0.206
set then equal and solve for T
10.3T + 51.5 = 1.256 + 6.68 + 0.0412T + 0.206
10.3T + 51.5 = 1.256 + 6.68 + 0.0412T + 0.206
10.3T + 51.5 = 8.142 + 0.0412T
10.3T – 0.0412T = 8.142 – 51.5
10.3T = -43.4
T = –4.2º
Then the heat required to warm the ice to 0º and to melt the ice and then to warm the ice water to T
is equal to the heat required to cool the water from 15ºC to T.
specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
to warm ice to T
E1 = 2.06 kJ/kgC x 0.2 kg x 5C
to melt ice
E2 = 334 kJ/kg x 0.2 kg
to warm water from 0C to T
E3 = 4.186 kJ/kgC x 0.2 kg x (T–0)
Total energy taken from the ice is
E4 = 2.06 kJ + 66.8 kJ + 0.837T = 68.86 kJ + 0.837T
to cool water
E5 = 4.186 kJ/kgC x 0.02 kg x (15–T)
E5 = 0.0837(15–T) = 1.2558 – 0.0837T
set them equal and solve for T
68.86 + 0.837T = 1.2558 – 0.0837T
0.837T + 0.0837T = 1.2558 – 68.86
this is negative, so not all the ice melts.
Starting over, there are two new possibilities.
1. the ice warms up to the melting point and some of it melts.
2. The ice doesn't warm up to the melting point.
In case 1, final temp is 0º
case 2, T<0º
to warm ice to T (a negative number)
change in temperature is T+5
E1 = 2.06 kJ/kgC x 0.2 kg x (T+5)
E1 = 10.3T + 51.5
To cool water to 0º
E2 = 4.186 kJ/kgC x 0.02 kg x 15 = 1.256 kJ
to freeze the water
E3 = 334 kJ/kg x 0.02 kg = 6.68 kJ
to cool water from 0 to T
E4 = 2.06 kJ/kgC x 0.02 kg x (T+5)
E4 = 0.0412T + 0.206
set then equal and solve for T
10.3T + 51.5 = 1.256 + 6.68 + 0.0412T + 0.206
10.3T + 51.5 = 1.256 + 6.68 + 0.0412T + 0.206
10.3T + 51.5 = 8.142 + 0.0412T
10.3T – 0.0412T = 8.142 – 51.5
10.3T = -43.4
T = –4.2º
-
I do have a question about how you solved it, when you did
"E1 = 2.06 kJ/kgC x 0.2 kg x (T+5)
E1 = 10.3T + 51.5" ; I do not get the same number you get using your numbers I get E1= 0.412T+ 2.06,
I would appreciate if you would tell me how you got the values you got?.. thank you! ;)
"E1 = 2.06 kJ/kgC x 0.2 kg x (T+5)
E1 = 10.3T + 51.5" ; I do not get the same number you get using your numbers I get E1= 0.412T+ 2.06,
I would appreciate if you would tell me how you got the values you got?.. thank you! ;)
Report Abuse
-
And once you solve for T with the answer I got, T would be 16.4 degC which would not make sense to be right...thank you for trying though, and since you seem to have a lot of knowledge on this you might be able to maybe go another way?
Report Abuse