Question

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Answer 1

Answer:

Explanation:

Solution:

If all the ice melted, first we find out how much heat it would require:

Heat abs by ice = heat of fusion = mHf =(20g)(80cal/g)(4.18J/cal)=6 688 J = ∆qice

Now, just to get an overall picture, let’s figure out how much heat would need to be removed from the water to cool it by the melting ice all the way down to 0°C:

Heat released by water to go to °C = 100(1cal/g°C)(4.18J/cal)(0-35°C) = - 14,630 J

OK since this is larger in magnitude than the heat that can be absorbed by the ice, we conclude that all the ice will melt to cool down the water. Furthermore, after all the ice has melted to 0°C water, we must take into account the fact that it will also need to equilibrate with the rest of the water to reach a final equilibrium temperature.

qice = -6688 J = qwater = mCp(Tf -35) = (100g)(1 cal/g°C)(4.18J/g°C)((T-35) solving for T:

-6688=418T–14630 => T=19°C

But this is not quite complete since we need to account for the 20 g of 0°C water that has just been formed from the melting of the ice to bring the 100 g of water down to 19°C. the 2 water temperatures must equilibrate:

So heat gained by the cold water + heat lost by the warmer water =0 m1C(T-0°C)+ m2C(T -19°C) =0 => (20)(T) + 100(T-19) =0

120T -1900 = 0 => T = 1900/120 = 15.8 °C final temperature.

After melting all the ice: ∆qw =- 6688 J = 100(4.18) ∆T =>∆T = 35-16= 17°C.