A piece of ice in water

1. Oct 17, 2014

Karol

1. The problem statement, all variables and given/known data
A 50 gr piece of ice at temp' -100C is thrown in water which is at 00C.
How much water will freeze around it. no heat is lost or gained from outside.

2. Relevant equations
Specific heat of ice: 0.55
Melting heat: 79.7[cal/gr/C0]

3. The attempt at a solution
The ice comes from -10C0 to 0C0 by taking heat from the water. this amount of heat freezes the water:
$50[gr]\cdot 10[C^0]=79.7\left[\frac{cal}{gr\cdot C^0}\right]\cdot m\rightarrow m=6.3[gr]$
Is it correct?

2. Oct 17, 2014

Staff: Mentor

Where did you take the specific heat of ice into account?
Also, your units do not match, but that has the same error as origin.

3. Oct 17, 2014

NTW

You have forgotten take into account the specific heat of ice: 0,55...

Solving it 'without equations', that supercooled piece of ice can absorb 50 * 10 * 0,55 = 275 cal from the outside without melting, and that 'heat debit' is used up in freezing some water that is already at 0º C, so that one gram of that water needs just 79,7 cal to freeze. You have 275 cal 'available for freezing'. Thus, you can freeze 275/79,7 = 3,45 g of water...

4. Oct 17, 2014

Karol

$0.55\left[\frac{cal}{gr\cdot C^0}\right]\cdot50[gr]\cdot 10[C^0]=79.7\left[\frac{cal}{gr}\right]\cdot m\rightarrow m=3.45[gr]$
Is it correct?

5. Oct 17, 2014

Staff: Mentor

Correct.

6. Oct 17, 2014

Thanks