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### Topic: Calorimetry  (Read 4943 times)

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#### Spoon-

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### Calorimetry
« on: November 13, 2007, 07:56:05 PM »
OK I've tried this question and the answer I'm getting doesn't match the one my teacher gave us.
Here's the question:

A 25.0g steel ball (c = 427J/(kg.K)) is quickly transfered from boiling water to a 200g calorimeter containing 50.0g of water at 25 degrees Celsius. If the final temperature is 27 degrees Celsius, calculate the specific heat of the calorimeter.

This is what I did:

(0.025kg)(427427J/kg.K)(27-100) = (0.05)(27-25)c
c = 7792.75 J/kg.K

The answer should be 902 J/kg.K. What am I doing wrong?

#### enahs

• 16-92-15-68 32-7-53-92-16
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• Gender:  ##### Re: Calorimetry
« Reply #1 on: November 13, 2007, 08:56:43 PM »
You are calculating the specific heat of water (wrongly).

Your system consists of three parts.
The water, the steel ball, and the calorimeter!

You might be familiar with the simple two system equation that you appear to be trying to use ( Q1=-Q2 )
However, it might be simpler to remember that is is actually written as.
Q1 + Q2 +...Qn = 0
As you are assuming there is no energy in the form of heat lost, nor is there any heat being added to the system by an outside source, right?

You know the Initial and final temperature of the metal ball, the water (and because every thing is in thermodynamic equilibrium, the calorimeter, 0th law of Thermodynamics) you just have three Q's = 0 (Q1 + Q2 + Q3 = 0)

You must also know the specific heat of water to work this question.

#### Spoon-

• Regular Member
•   • Posts: 11
• Mole Snacks: +0/-0 ##### Re: Calorimetry
« Reply #2 on: November 13, 2007, 10:42:42 PM »
You are calculating the specific heat of water (wrongly).

Your system consists of three parts.
The water, the steel ball, and the calorimeter!

You might be familiar with the simple two system equation that you appear to be trying to use ( Q1=-Q2 )
However, it might be simpler to remember that is is actually written as.
Q1 + Q2 +...Qn = 0
As you are assuming there is no energy in the form of heat lost, nor is there any heat being added to the system by an outside source, right?

You know the Initial and final temperature of the metal ball, the water (and because every thing is in thermodynamic equilibrium, the calorimeter, 0th law of Thermodynamics) you just have three Q's = 0 (Q1 + Q2 + Q3 = 0)

You must also know the specific heat of water to work this question.

Oh... I get it. Thanks, lol can't believe I didn't see that.