How much water will freeze when a piece of ice is thrown into it?

[Karol]

Homework Statement

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.

Homework Equations

Specific heat of ice: 0.55
Melting heat: 79.7[cal/gr/C⁰]

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?

 

[mfb]

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.

 

[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...

 

[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?

 

[mfb]

Correct.

 

[Karol]

Thanks