How to Calculate Energy Released in an Aluminum Calorimeter Reaction?

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SUMMARY

The energy released in the aluminum calorimeter reaction is calculated using the formula q=mcΔT. In this scenario, with 100g of water and a 50g aluminum calorimeter, the temperature change from 20°C to 25°C results in a total energy release of 2317.5J. The specific heat capacities used are 4.18 J/g°C for water and 0.91 J/g°C for aluminum. The calculation can be simplified by summing the heat absorbed by both substances rather than using proportions.

PREREQUISITES
  • Understanding of specific heat capacity
  • Familiarity with the formula q=mcΔT
  • Basic knowledge of exothermic reactions
  • Ability to perform unit conversions and calculations
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  • Learn detailed applications of the specific heat capacity formula in different materials
  • Explore the concept of calorimetry and its practical uses
  • Investigate the principles of energy conservation in chemical reactions
  • Study advanced thermodynamics related to heat transfer and reaction kinetics
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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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