Specific heat problem help needed.

[Emi]

100g of a substance with a Specific heat of 0.1 cal/gCelcius put into a calorimeter containing 200g of water. The initial temp. of the substance was 250 degrees celcius and the initial temp. of the water is 0 degrees celcius. What will the final temp. of the mixture be? (ignore heat gained by calorimeter)

[swansont]

100g of a substance with a Specific heat of 0.1 cal/gCelcius put into a calorimeter containing 200g of water. The initial temp. of the substance was 250 degrees celcius and the initial temp. of the water is 0 degrees celcius. What will the final temp. of the mixture be? (ignore heat gained by calorimeter)

 

What have you got so far? Equations you might use?

 

Hint: energy is conserved.

[Emi]

calories= (mass) x (change in temperature) x (specific heat) Do you use this formula for the substance and then for the water and add the calories together? I can't figure what the change in temperature would be for either equation.

[swansont]

Any heat lost by the object is gained by the water. You also know that the final temperatures have to be equal.

 

The one potential stumbling block is if there is a phase change in the water, so you have to check for that.

[Emi]

Would 250C be the change in substance temperature and the amount of calories for the substance change in temperature then be used to figure the change in water temperature which would then be the final temperature of both?

[Karnage]

(mass of substance)(specific heat of substance)(change in temp) = (mass of water)(specific heat of water)(change in temp)

 

You use this equation becuase energy is conserved. The change in temp of the substance will be (250-x) where x is the final temperature. The change in temp of water will be 0+x or simply x. (again x denotes the final temperature of the mixture)

 

(100g)(.1)(250-x)=(200g)(1)(x)

 

if you do it right, x should equal about 11.9 degrees celsius. This I think is the final temperature of the whole thing. My chemistry is a bit rusty (I am more of an expert in biology and biochemistry) and i havent done it in two years. However i think this is the right answer to your problem. If any1 thinks otherwise please place a reply.

[swansont]

(mass of substance)(specific heat of substance)(change in temp) = (mass of water)(specific heat of water)(change in temp)

 

You use this equation becuase energy is conserved. The change in temp of the substance will be (250-x) where x is the final temperature. The change in temp of water will be 0+x or simply x. (again x denotes the final temperature of the mixture)

 

(100g)(.1)(250-x)=(200g)(1)(x)

 

if you do it right' date=' x should equal about 11.9 degrees celsius. This I think is the final temperature of the whole thing. My chemistry is a bit rusty (I am more of an expert in biology and biochemistry) and i havent done it in two years. However i think this is the right answer to your problem. If any1 thinks otherwise please place a reply.[/quote']

 

The point here isn't to do the homework for the original poster, it's to get him or her to be able to do it, and actually learn something.

[Emi]

Thanks for everyone's help. I was stuck because of missing a lecture.

[Karnage]

Sorry dude i didnt know im kinda new to this forum. But i hope my explanation gives the guy an insight. You know, sometimes telling the answer can make the person realize the method how to do it and apply it to other problems you know...

[LazerFazer]

Yup, I agree. Showing someone how its done usually helps them see a pattern and then they can figure out future problems themselves.

[swansont]

Yup, I agree. Showing someone how its done usually helps them see a pattern and then they can figure out future problems themselves.

 

Yes, but one has to expect that an example or two were worked out in the textbook. We don't want to become enablers for a generation of homework addicts.