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Thermal physics verification

so this topic has been quite hard for me so far despite taking chemistry too but I just want to verify that I got the correct answer, if someone doesn't mind working it out:

0.38kg of unknown liquid at 12 Celsius is heated by copper can weighing 0.9kg by a 20W heater for 3 mins. "It" (im guessing the liquid) reaches 17 celsius. find the specific heat capacity of the liquid.
My working out:
since E=Pt and we take the liquids details:
mcT=Pt
0.38 x c x (17-12)= 20 x (3x60)
1.9c=3600
c=1894.7 (1dp)
Original post by Toasticide
so this topic has been quite hard for me so far despite taking chemistry too but I just want to verify that I got the correct answer, if someone doesn't mind working it out:

0.38kg of unknown liquid at 12 Celsius is heated by copper can weighing 0.9kg by a 20W heater for 3 mins. "It" (im guessing the liquid) reaches 17 celsius. find the specific heat capacity of the liquid.
My working out:
since E=Pt and we take the liquids details:
mcT=Pt
0.38 x c x (17-12)= 20 x (3x60)
1.9c=3600
c=1894.7 (1dp)


This isn't a brilliantly worded question but the fact that they've mentioned the weight of the copper can makes me think they want you to take that into account too (because obviously the heat isn't just going to heat the liquid, it will heat the can too). Which is slightly problematic because you have to assume the can started at the same temperature as the liquid (and reaches the same end temperature).

If you ignore the copper can then your working is correct, but I don't think they want you to ignore it.
Reply 2
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Toasticide
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Original post by Plagioclase
This isn't a brilliantly worded question but the fact that they've mentioned the weight of the copper can makes me think they want you to take that into account too (because obviously the heat isn't just going to heat the liquid, it will heat the can too). Which is slightly problematic because you have to assume the can started at the same temperature as the liquid (and reaches the same end temperature).

If you ignore the copper can then your working is correct, but I don't think they want you to ignore it.


Yeh they didn't mention any temperatures for the copper. been trying this a while now, can't form simultaneous equations with both variables as you can only obtain one equation (energy released by copper=that taken in by liquid=Pt)
Thanks for verifying the other stuff though :smile:
Original post by Toasticide
Yeh they didn't mention any temperatures for the copper. been trying this a while now, can't form simultaneous equations with both variables as you can only obtain one equation (energy released by copper=that taken in by liquid=Pt)
Thanks for verifying the other stuff though :smile:


You need to split this problem into parts.

You've got one part of the problem solved already, you know exactly how much energy is being put into the system (3 x 60 x 20 = 3600J). We also know exactly how much energy is being used to heat the copper because there are no unknowns in that equation (we can look up the SHC which is 385 J/K/kg, we assume that the change in temperature is 5K and we know the mass is 0.9kg).So you can calculate how much energy has been used to heat the copper, and therefore how much energy has been used to heat the liquid, and from that solve for the liquid's SHC.
Reply 4
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Toasticide
OP
17
Original post by Plagioclase
You need to split this problem into parts.

You've got one part of the problem solved already, you know exactly how much energy is being put into the system (3 x 60 x 20 = 3600J). We also know exactly how much energy is being used to heat the copper because there are no unknowns in that equation (we can look up the SHC which is 385 J/K/kg, we assume that the change in temperature is 5K and we know the mass is 0.9kg).So you can calculate how much energy has been used to heat the copper, and therefore how much energy has been used to heat the liquid, and from that solve for the liquid's SHC.


Ah thanks so much :smile:

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