aqa a2 specimen paper question
http://filestore.aqa.org.uk/resources/chemistry/AQA-74051-SQP.PDF
In question 9.1, i dont understand what the question is actually asking?
and why in the mark scheme, it said extrapolate back to time = 0 and determine the temperature? (Surely it would be 20 degrees?)
Thanks.
In question 9.1, i dont understand what the question is actually asking?
and why in the mark scheme, it said extrapolate back to time = 0 and determine the temperature? (Surely it would be 20 degrees?)
Thanks.
When the KCl dissolves, the temp of the solution drops and heat starts to flow into the solution from the surroundings.
The MS no doubt explains the method for a cooling curve (or in this case a heating curve). In which you assume a fairly fast reaction (in this case dissolution) and heat gain from the surroundings slowly returning the temp to that of the room. You extrapolate the curve back to the point of addition of the KCl to the water to see what temp it would have achieved had the dissolution actually been really fast and the temp had changed immediately.
The MS no doubt explains the method for a cooling curve (or in this case a heating curve). In which you assume a fairly fast reaction (in this case dissolution) and heat gain from the surroundings slowly returning the temp to that of the room. You extrapolate the curve back to the point of addition of the KCl to the water to see what temp it would have achieved had the dissolution actually been really fast and the temp had changed immediately.
Original post by Pigster
When the KCl dissolves, the temp of the solution drops and heat starts to flow into the solution from the surroundings.
The MS no doubt explains the method for a cooling curve (or in this case a heating curve). In which you assume a fairly fast reaction (in this case dissolution) and heat gain from the surroundings slowly returning the temp to that of the room. You extrapolate the curve back to the point of addition of the KCl to the water to see what temp it would have achieved had the dissolution actually been really fast and the temp had changed immediately.
The MS no doubt explains the method for a cooling curve (or in this case a heating curve). In which you assume a fairly fast reaction (in this case dissolution) and heat gain from the surroundings slowly returning the temp to that of the room. You extrapolate the curve back to the point of addition of the KCl to the water to see what temp it would have achieved had the dissolution actually been really fast and the temp had changed immediately.
isnt when KCl is is being added to the solution, the solution is at 20 degrees?
Original post by Pigster
time = 0 refers to when the KCl is added and theoretically, when the dissolution is complete, not when the experiment starts.
Thankyou for the explanation, but i am still confused on how to extrapolate to t = 0? How does the graph should look like?
Your graph is fine, essentially what you have to do is imagine the upwards line on the right is straight and extrapolate to t = 0
You should have done something like the following:
0 min 20.0 oC
1 min 20.5
2 min 20.0 (therefore average = 20.2)
3 min no reading (KCl added)
4 min 10.0 (from now on they will be all nice whole numbers to make the pattern easier to spot)
5 min 11.0
6 min 12.0
7 min 13.0
8 min 14.0
9 min 15.0
If you plotted that, you'd see that it was getting hotter as heat flowed from the surroundings.
What would the temperature been at t = 3 min? Extrapolation suggests 9. oC, Delta T = 20.2 - 9.0 = 11.2 oC
You should have done something like the following:
0 min 20.0 oC
1 min 20.5
2 min 20.0 (therefore average = 20.2)
3 min no reading (KCl added)
4 min 10.0 (from now on they will be all nice whole numbers to make the pattern easier to spot)
5 min 11.0
6 min 12.0
7 min 13.0
8 min 14.0
9 min 15.0
If you plotted that, you'd see that it was getting hotter as heat flowed from the surroundings.
What would the temperature been at t = 3 min? Extrapolation suggests 9. oC, Delta T = 20.2 - 9.0 = 11.2 oC
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