a- Calculate the heat, necessary to eliminated 200 g of liquid water from a cont

Question

a- Calculate the heat, necessary to eliminated 200 g of liquid water from a container which has 400 g of ice at -10 C0.

b- Calculate the specific heat of an unknown material with mass equal to 65 g if during an experiment the unknown sample at 100 oC, was put into a calorimeter–water system at room temperature (24 oC) and the final temperature (unknown material, calorimeter and water ) after few minutes was 85 oF. (Consider the mass of the calorimeter equal to 20 g and the water 100 g and the specific heat of the calorimeter 0.9 J/g.C and the specific heat of the water 4.186 J/g.C).

Explanation / Answer

As an example, suppose we mixed 50.0 grams of hot water at 60.0oC with 100.0 grams of cold water at 30.0oC.

From the equation above, we calculate:

tf = (50.0 x 60.0 + 100.0 x 30.0)/(50.0 + 100.0)
tf = (3000 + 3000)/(150.0)
tf = 40.0 oC

Assuming that all of the ice melts, the final temperature will be XoC.

1. 10.0 g ice at 0.0 oC melts to water at 0.0oC:
  H1 = (10.0 g)(334 J/g) = 3340  J

2. 10.0 g water at 0.0 oC warms to the final temperature:
  H2 = (10.0 g)(4.18 J/g-K)(X - 0)K = 41.8 X  J

3. 100.0 g water at 30.0 oC cools to the final temperature:
  H3 = (100.0 g)(4.18 J/g-K)(X - 30.00)K = 418(X - 30.0)  J

4. The apparatus at 30.0 oC cools to the final temperature:
  H4 = (10.0 J/K)(X - 30.00)K = 10.0(X - 30.0)  J
The net change in enthalpy is zero:
  H1 + H2 + H3 + H4 = 0

    3340 + 41.8 X + 418(X - 30.0) + 10.0(X - 30.0) = 0

    (41.8 + 418 + 10.0) X = (418)(30.0) + (10.0)(30.0) - 3340

        417 X = 9500

              X = 20.2 oC.

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