Relevant long, hard lab questions dealing with acid-base neutralization (part 3)?
cup. How would the measured value have been affected?
3. In the calorimetry lab how would your results have been affected if you had not used a lid to cover the
polystyrene cup after the acid and base solutions had been mixed? What would have been the effect on the
measurement of temperature and why?
5. A calorimeter contains 25.00 g water at 23.8 0C. A 5.00 g hot metal at 78.3 0C dropped into the calorimeter.
Calculate the specific heat of unknown metal if the final temperature of mixture is found to be 46.4 0C.
6. Consider a mixture of 75.00 g water at room temperature 23.0 0C and 25.0 g water at 95.0 0C.
Calculate the equilibrium temperature of the mixture.
Answer:
1) Well, glass is a cooled crystal between a crystalline and a liquid. It will not insulate heat as good as the styrofoam cup. The relative error would have been greater.
3) Well, the temperature in heat form would have been released giving you an in-accurate reading of the heat and temperature loss.
5) C = m x q x deltaT
C = 5g x q 31.9 You don't know the heat released. You should have a table with possible "heats" and you could plug in each value to find a possible value for Specific heat. :)
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