Calculating the final temperature of a mixture of Ice and Water

[Jham808]

Homework Statement

How do I calculate the following. The final temperature of a mixture of of ice and water. Where 176grams of ice at -10celsius mixes with 206 grams of water at 73.3celsius. I have tried this equation in multiple fashions and cannot seems to come to a consistent answer! Any help would be appreciated!

Homework Equations

The Attempt at a Solution

Why does the below not work?
water specific heat=4.184
ice specific heat= 2.11
heat of fusion=334

((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))+((specific Heat ice)X(Mass ice)x(x-0))=((specific Heat water)X(Mass water)x(∆temperature))

 

[ehild]

Homework Statement

How do I calculate the following. The final temperature of a mixture of of ice and water. Where 176grams of ice at -10celsius mixes with 206 grams of water at 73.3celsius. I have tried this equation in multiple fashions and cannot seems to come to a consistent answer! Any help would be appreciated!

Homework Equations

The Attempt at a Solution

Why does the below not work?
water specific heat=4.184
ice specific heat= 2.11
heat of fusion=334

((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))+((specific Heat ice)X(Mass ice)x(x-0))=((specific Heat water)X(Mass water)x(∆temperature))

What do you mean on ∆temperature?

After the ice reached the melting temperature and melted, it became water at 0 degree, and its specific heat is the same as that of water.

ehild

 

[Jham808]

Hello ehlid,

I have tried this in stages but continue to do something incorrectly.

The Amount of Energy in Water
((specific Heat water)X(Mass water)x(∆temperature))
4.18 X 73.3C X 206gram=63,117Joules

The Amount of Energy to get ice to melting point
((specific Heat ice)X(Mass ice)x(∆temperature))
2.11 * 10c * 176grams=3,713Joules

Energy for Melting
((mass ice)*(heat of fusion))
334*176=58,784

Total Energy of ice melt and bringing to melting point
((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))
62,497

62,497<63,117 complete melting of ice occurs

the remaining energy is 602 joules
the total remaining water mass is 176+206=326grams
602joules=326grams * 4.18 * (x-0)
602= 1,362 * (x-0)
0.44=(x-0)

is this correct?

 

[lurflurf]

^mostly
remaining energy=
-62,497+63,117=620 not 602
typo?

 

[Jham808]

yes, 620 is correct. apologies. Is this the correct answer?

 

[ehild]

I have tried this in stages but continue to do something incorrectly.

The Amount of Energy in Water
((specific Heat water)X(Mass water)x(∆temperature))
4.18 X 73.3C X 206gram=63,117Joules

Why did you ignore the last digit of the specific heat? 4.184 X 73.3C X 206gram=63,522Joules

The Amount of Energy to get ice to melting point
((specific Heat ice)X(Mass ice)x(∆temperature))
2.11 * 10c * 176grams=3,713Joules

Energy for Melting
((mass ice)*(heat of fusion))
334*176=58,784

Total Energy of ice melt and bringing to melting point
((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))
62,497

62,497<63,117 complete melting of ice occurs

the remaining energy is 602 joules

It is 1029 J

the total remaining water mass is 176+206=326grams
602joules=326grams * 4.18 * (x-0)
602= 1,362 * (x-0)
0.44=(x-0)

is this correct?

No. Recalculate with the correct remaining energy. But the method is correct.

ehild

 

[lurflurf]

I wonder if the desired answer should be 0°C.
Can you quote the question exactly?
It seems we are to use significant 3 digits for each value.
Then we should write -10.0°C.
When we subtract our 3 digits will be reduced to 1.
This leads to an answer of about 0.4°C.
As this value is uncertain the mixture might contain some unmelted ice or be slightly warmer than 0°C. Not that it would make much difference.