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If 142.38 g of l water at 21.7 degrees C is placed into a syrofoam cup with 174.36 g of ice at -27.5 degrees C what would be the final temperature of the entire contents at equilibrium? If it is partially frozen, how many g of the ice has melted?

given values
cp water = 4.184 j/g degree c
cp ice = 2.114 j/g degree c
heat of fusion of ice is 335 J/g

so far, i have this:

(174.36g)(2.114g)(27.5 C)
= 10247 g to three sig figs = 10200 J

is that i calculate the energy required to raise the ice to 0 degrees C?

now i know that the next steps are to find the enrgy required to melt the ice and the energy reuired to cool the water down to 0 degrees C. how do i do those two calculations?

after that i did this to try and melt the ice
q = (mass l H20/molar mass H20)X(335 j/g) the 335 is the heat of fusion.

is that correct? if not, what am i supposed to do?

  • Jake - ,

    Your 10,247 J I get 10,136 J. Check my arithmetic. THEN, I would keep all the numbers and round at the end; otherwise, it may lead to rounding errors. I'll assume the 10,136 J is correct to move the ice from -27.5 to Oo C.

    Then we need to calculate the energy required to cool the liquid water from 21.7 C to zero C. That is
    142.38 x 4.184 x 21.7 = check me on this but I get about 13,000 J or so. You do the exact number.

    So if you look at the numbers, we can move the ice to zero and move the lqiuid water to zero and have about 3,000 J left in the liquid water. That will melt some of the ice. How much?
    q = mass x 335 J/g. Calculate the mass of ice that will melt.

    I hope this helps. Check my thinking. Check my work.

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