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calculating temperature change

It is possible to use the equations below to predict the temperature drop of the water when a chosen amount of ice is added to it. Will the actual temperature drop, measured in your experiment, be equal to the predicted value? Use relevant scientific explanations in your answer.Use the results of your experiment, appropriate calculations and your research (Part 1) to provide evidence to support your answer.

1 cm3 of water has a mass of 1 g
energy = mass × specific heat capacity × temperature change
energy = mass × specific latent heat

How would I answer the above question? And if it helps:
In my experiment I used 200ml, amount of ice was 15g. Specific latent heat capacity of fusion is 334 KJ/Kg, specific heat capacity 4.18 KJ/Kg.
(edited 10 years ago)
Reply 1
Avatar for Borek
Borek
4
First, ice melts (how much heat would it need?), and warm water supplies the heat to melt it (how much heat did the water supply?).

Then, cold water from the melted ice gets heated by the warm water, till their temperatures are identical (heat gained = heat lost).

Can you write the heat balance?
Reply 2
Avatar for tammie94
tammie94
OP
15
Original post by Borek
First, ice melts (how much heat would it need?), and warm water supplies the heat to melt it (how much heat did the water supply?).

Then, cold water from the melted ice gets heated by the warm water, till their temperatures are identical (heat gained = heat lost).

Can you write the heat balance?


I understand that part but how do I work it out using the information they gave?
Reply 3
Avatar for Borek
Borek
4
I am not sure what the problem is.

Do you have experimental values? If so, calculate the result and see if it agrees with theoretical prediction. If not, all you can say is that the experimental drop should be reasonably close to the calculated one. It won't ever be exact, because of experimental errors.
Reply 4
Avatar for tammie94
tammie94
OP
15
Original post by Borek
I am not sure what the problem is.

Do you have experimental values? If so, calculate the result and see if it agrees with theoretical prediction. If not, all you can say is that the experimental drop should be reasonably close to the calculated one. It won't ever be exact, because of experimental errors.


This is how I worked it out:

energy = mass × specific latent heat
= 0.015 x 334 = 5.01 kJ

energy = mass × specific heat capacity × temperature change
temperature change = energy/ mass x specific heat capacity = 5.01/ 0.015 x 4.18 = 80 degrees, but how can the temperature change be so big, in my experiment the temperature change was 5 degrees:confused:
Reply 5
Avatar for tammie94
tammie94
OP
15
anyone :please:
Reply 6
Avatar for Borek
Borek
4
Original post by tammie94
temperature change = energy/ mass x specific heat capacity = 5.01/ 0.015 x 4.18 = 80 degrees, but how can the temperature change be so big, in my experiment the temperature change was 5 degrees:confused:


0.015 kg is not mass of the water that was cooling down.
Reply 7
Avatar for tammie94
tammie94
OP
15
Original post by Borek
0.015 kg is not mass of the water that was cooling down.


I managed to work it out, I got -5.99. Thank you so much for the help! :smile:
Reply 8
Avatar for Borek
Borek
4
Original post by tammie94
I managed to work it out, I got -5.99. Thank you so much for the help! :smile:


Next time list everything in your first post. We have lost a lot of time because you have failed to post all relevant information and I had to poke blindly not knowing where the problem is.

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