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Topic: Specific Heat Capacity  (Read 4420 times)

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Offline yooper_at_heart678

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Specific Heat Capacity
« on: October 09, 2007, 06:44:11 PM »
I am confused and although I was talked through a problem with the teacher I'm still lost.
The first problem I think I did right but I'm confused with rounding it to the right amount of  significant figures

As the detective investigates a fire he notices that several windows have actually melted. He hypothosizes that fire changed the temperature of the room from room temperature(22.0 C) to at least 80 C. How much energy was added to a 6.00 kg window.(The specific heat if glass is 0.502)
6000X .502 j/ g* degrees CX58.0= 174696(How would you round that? Without a decimal point its confusing)

To raise a temperature of 50.0 ml of water from 25 degrees C to the boiling point how much heat do we have to add?(The density of water is 1.00 g/ml The specific heat is 4.184 J/g degrees C)


If.1570 J of heat is added to a 50.0 g metal bar and its temp changes from 20.0 degrees C to 55.0 What metal is it?


Can someone walk me through these?

Offline Padfoot

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Re: Specific Heat Capacity
« Reply #1 on: October 10, 2007, 01:16:28 AM »
To raise a temperature of 50.0 ml of water from 25 degrees C to the boiling point how much heat do we have to add?(The density of water is 1.00 g/ml The specific heat is 4.184 J/g degrees C)


If.1570 J of heat is added to a 50.0 g metal bar and its temp changes from 20.0 degrees C to 55.0 What metal is it?


Can someone walk me through these?

These questions are exactly the same as Q1, you are only solving for a different variable.  Rearrange your eqn.
For Q3 first calculate specific heat of the metal.

Offline yooper_at_heart678

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Re: Specific Heat Capacity
« Reply #2 on: October 10, 2007, 05:17:21 PM »
.
To raise a temperature of 50.0 ml of water from 25 degrees C to the boiling point how much heat do we have to add?(The density of water is 1.00 g/ml The specific heat is 4.184 J/g degrees C)


Ok so 4.174 J/g cx50 gx50 C? I have a test on this material tommorow so I need all the help I can get

Offline Padfoot

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Re: Specific Heat Capacity
« Reply #3 on: October 10, 2007, 06:07:29 PM »
This question is exactly the same as Q1! 
Apologies for my last post, it may have confuddled you - my comment should only relate to Q3.

m=50g (50mL*1gmL-1)
changeT= 100-25=75

E=mC*changeT
  = (50*4.18*75)J

Start point for Q3:
E=mC*changeT
C=E/(m*changeT)

Hope that helps  :)

Offline yooper_at_heart678

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Re: Specific Heat Capacity
« Reply #4 on: October 10, 2007, 09:50:04 PM »
Yes it does thank you so much!

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