How to Calculate Energy Released in an Aluminum Calorimeter Reaction?

[hamza2095]

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?

 

[Borek]

Why do you complicate things instead of calculating separately heat absorbed by aluminum and heat absorbed by water and then summing it?

 

[hamza2095]

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

 

[Borek]

The simplest thing to do:

$q=m_{H_2O}c_{H_2O}\Delta T + m_{Al}c_{Al}\Delta T = (m_{H_2O}c_{H_2O} + m_{Al}c_{Al})\Delta T$

 

[hamza2095]

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.