How Do You Calculate Heat Capacity and Specific Heat in Chemistry Problems?

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Discussion Overview

The discussion revolves around calculating heat capacity and specific heat in chemistry problems, specifically focusing on two example problems involving aluminum and an organic compound combusted in a calorimeter. Participants are examining the application of the formula q=mcT and the definitions of heat capacity and specific heat.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving a 2kg sample of aluminum absorbing 64.4 kJ of heat and calculates the temperature increase using q=mcT, arriving at a different answer than expected.
  • Another participant questions the calculation, asking for clarification on the conversion of heat capacity (C) to specific heat (c) and the mass used in the calculations.
  • There is confusion regarding the application of the heat capacity of the calorimeter versus the specific heat of the substance being analyzed.
  • Participants discuss the need for clarity on the definitions of heat capacity and specific heat, as well as the proper use of the given mass in calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations presented. There are multiple viewpoints regarding the correct application of formulas and definitions, and the discussion remains unresolved.

Contextual Notes

There are limitations in the clarity of the calculations and definitions used by participants, particularly regarding the conversion between heat capacity and specific heat, as well as the proper mass to use in the calculations.

luysion
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hey for this question
If a 2kg sample of aluminium absorbs 64.4 kJ of heat from its surrounds by how much does its temperature increase

the answer is 65.1

how? i got 32.5
q=mcT
T=q/mc=(64.6x10^3)/(2000x.9930)

also same for this question
If 0.5g of an organic compound is completely combusted in air in a constant press calorimeter with a heat capacity of 510 J K-1 a temperature rise of 5.9 degrees is recorded. The heat of combustion (kJ g-) of the compound is approximately

the answer is 3009...
i convered C to c then used q=mcT
which gave me 3009 J , so shouldn't it be 3.00 kJ
so 3.00/0.5 to get kJ per gram
 
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To some extent you are twice right (that is assuming specific heat of Al is 0.9930). Although details are fuzzy at best.

Can you elaborate on

luysion said:
i convered C to c

You mean you converted 5.9 to 5.9?

luysion said:
then used q=mcT

And what did you put for m?
 
Yeah where did you get the mass lol.
 
Cheers for the reply guys,
Um, i meant C as in heat capacity and c as in specific heat (c=C/m)
the mass was given as 2kg
 
luysion said:
Um, i meant C as in heat capacity and c as in specific heat (c=C/m)

Show precisely what you did and how, as as far as I can tell you did something else. You should use heat capacity of the calorimeter to calculate amount of heat (and numbers show that's what you did), not specific heat. And if you calculated specific heat - specific heat of what, and what have you used for mass (c=C/m)?
 

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