Adding Ice to Water Problem

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  • #1
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Question:

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9C

How much ice at a temperature of −10.0C must be dropped into the water so that the final temperature of the system will be 26.7C?

Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.


Working out:

Q_b = m_beaker * c_water * change in temp. of beaker
Q_b = (0.205)(4190)(26.7-79.9) = -45696.14J

Q_1 = m_ice * c_ice * change in temp. of ice
Q_1 = m_ice(2100)(26.7- -10) = m_ice(77070)

Q_2 = m_ice * L_f
Q_2 = m_ice(334000)

Q_b + Q1 + Q2 = -45696.14 + m_ice(77070) + m_ice(334000) = 0

Solving this for m_ice, I get: 0.111kg


I'm told that the answer is wrong with a hint saying "Remember that the ice turns into liquid water above the melting point."


I'm not sure how to go about this, can someone help me out, isn't that hint above the working out I did for Q_2 ?
 

Answers and Replies

  • #2
learningphysics
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Logistics said:
Q_1 = m_ice * c_ice * change in temp. of ice
Q_1 = m_ice(2100)(26.7- -10) = m_ice(77070)

The mistake is above I think... the ice first heats up... then turns to water then heats up again... so it should be
Q_1 = m_ice(2100)(0--10) + m_ice(4190)(26.7-0)= 132873 m_ice

Q_2 and Q_b look right.
 
  • #3
dextercioby
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You analysis is incorrect.

1.The ice is will receive heat in the domain -10°--->0°.
2.The ice at 0° will receive heat to pass into liquid water (no temperature & pressure variation).
3.The initial mass of ice,now in the liquid state will receive heat to reach equilibrium temp.
4.The initial mass of hot water will give away heat (all the heat added from points 1.-3.) till it reaches equilibrium temp.

These hints are all u need...For now.

Daniel.
 
  • #4
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dextercioby I understand the theory behind it, I have a nice diagram that explains this. But I don't know how to put this into numbers

At the time I posted my problem I have thought of what learningphysics said and got
Q_1 = m_ice(2100)(0--10) + m_ice(4190)(26.7-0)= 132873 m_ice

But I'm not entirely sure that it is correct
 
  • #5
learningphysics
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Logistics said:
dextercioby I understand the theory behind it, I have a nice diagram that explains this. But I don't know how to put this into numbers

At the time I posted my problem I have thought of what learningphysics said and got
Q_1 = m_ice(2100)(0--10) + m_ice(4190)(26.7-0)= 132873 m_ice

But I'm not entirely sure that it is correct

If you replace your Q1 with the one above... I think you'll get the right answer...

I'd organize it like this:

Heat change of the water already in the beaker:
Q_b = (0.205)(4190)(26.7-79.9) = -45696.14J (just as you did)


Net het change of the ice as it turns to water at 26.7 (3 parts) :
Heat change of the ice while it is still in ice form (it's temperature is rising from -10 to 0):
1)Q_i = m_ice(2100)(0--10)

Heat change of the ice as it turns to liquid (temperature is at 0):
2)Q_c = m_ice(334000)

Heat change of the liquid that was previously ice (temperature rises from 0 to 26.7):
3)Q_d = m_ice(4190)(26.7 - 0)

Q_ice = Q_i + Q_c + Q_d


So total heat change = Q_b+ Q_ice =0
 
Last edited:
  • #6
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Thanks for your help guys.

I worked it out, what you said learningphysics, I done it before you posted just wasn't sure if it was right. Didn't think it was that simple. So didn't bother putting it in my 1st post :p


Thanks again.
 

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