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A 250g piece of ice at 0.0 degrees C is placed in 200g of water at 100 degrees C. What is the final temp.?

Heat of fusion of ice is 335 J/g

Specific heat of ice is 2.1 J/g degrees C

Specific Heat of water is 4.2 J/ g degrees C

Possible answers:

a) 0.0 degrees

b) 2.5 degrees

c) 10 degrees

d) 45 degrees

3 Answers

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    Specific heat of ice is unnecessary for this question and (although it probably won't make a difference), I think the currently accepted values for heat of fusion and specific heat of water are 334 J/g and 4.18 J/g degrees C.

    The first step is to find the amount of heat absorbed by the ice in order to melt. This is simply number of grams*heat of fusion, which is 83500 joules. Then, use the formula q=mCdeltaT, in which q is heat, m is mass, C is specific heat, and delta T is the change in temperature.

    83500=200(4.18)x

    83500=836x

    99.88=x

    I guess I should have used the values you gave, but 99.88 is close enough to 100 that we can reasonably assume that's what the answer would have come out to. Since the change in temperature was 100 degrees C and it decreased, 100 degrees C-100 degrees C=0 degrees C

    a) 0.0 degrees

  • 1 decade ago

    Lets say the ice is going to melt and reach a temperature of X

    The the energy required (J) by the ice first melting and then reaching that temperature will be

    250 x 335 + 250 x 4.2 (X-0) = 83750 + 1050X

    The energy released (J) by the hot water cooling to X will be

    200 x (100 - X) x 4.2 = 84000 - 840X

    Now these 2 quantities will be equivalent so

    83750 + 1050X = 84000 - 840X

    and solving gives...

    1890X = 250

    X=0.13 deg C

    so I'd also choose answer a.

  • 1 decade ago

    at 0 C, water can be either solid or liquid, so no heating needs to occur to melt the ice. The ice will absorb 83750 J in order to melt.

    The water has to give the energy, so it would have to drop in temperature. dividing the total energy lost (83750) by the number of grams (200) gives us the amount of energy lost by each gram of water (418.75 J). Divide this amount by the J/g per C degree of water (4.2) and we get the final temperature drop of the water (99.70).

    As we can see, the water is very close to 0 degrees (0.3) before the ice has melted, so the final temperature of the water and melted ice will be between 0.0 and 0.3, or effectively 0.0.

    The answer is most likely A.

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