How to Calculate Energy Released in an Aluminum Calorimeter Reaction?

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Discussion Overview

The discussion revolves around calculating the energy released in an exothermic reaction involving a 100g water sample and a 50g aluminum calorimeter, with a temperature change from 20 degrees Celsius to 25 degrees Celsius. Participants explore different methods for calculating the heat transfer using specific heat capacities.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes using proportions based on the total mass and specific heat capacities to calculate the energy released, resulting in a total of 2317.5J.
  • Another participant questions the initial approach, suggesting that calculating the heat absorbed by aluminum and water separately and then summing them would be simpler and potentially yield the same result.
  • A later reply reiterates the suggestion to calculate separately, indicating uncertainty about the initial method's complexity.
  • Another participant provides a formula that combines the heat absorbed by both substances into a single equation, asserting that it simplifies the calculation while confirming it yields the same answer as the initial approach.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the initial approach versus the proposed separate calculations. There is no consensus on the best method, as some prefer the initial method while others advocate for the separate calculations.

Contextual Notes

Participants do not clarify whether any assumptions about heat loss to the environment or other factors are considered in their calculations.

hamza2095
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Homework Statement


There is 100g of water inside of a 50g aluminum calorimeter. After a reaction takes place the temperature of the calorimeter goes from 20 degrees celsius to 25 degrees celsius

Find the energy released by this exothermic reaction

specific heat capacity of water is 4.18 and specific heat capacity of the aluminum calorimeter is 0.91

Homework Equations


q=mcT

The Attempt at a Solution


I approached this question using proportions since the total mass is 150, 1/3rd of the specific heat capacity is that of aluminum and 2/3 is that of water.

q=150g(2(4.18)/3 + 0.91/3)(25-20)
=2317.5J

Is this approach correct?
 
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Why do you complicate things instead of calculating separately heat absorbed by aluminum and heat absorbed by water and then summing it?
 
Borek said:
Why do you complicate things instead of calculating separately heat absorbed by aluminum and heat absorbed by water and then summing it?
Would that not also give the same answer due to bedmas? I thought of doing this but wasn't sure how to
 
The simplest thing to do:

q=m_{H_2O}c_{H_2O}\Delta T + m_{Al}c_{Al}\Delta T = (m_{H_2O}c_{H_2O} + m_{Al}c_{Al})\Delta T
 
Borek said:
The simplest thing to do:

q=m_{H_2O}c_{H_2O}\Delta T + m_{Al}c_{Al}\Delta T = (m_{H_2O}c_{H_2O} + m_{Al}c_{Al})\Delta T
Thank you! It also gives the same answer as I got but its good to see it in simpler form.
 

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