Help with chem problem?
calculate the specific heat of the metal in joules/g degrees celcius.
anything to help me to set it up.
Answer:
I'm a college sophomore- it's been a while since chem, so you may want to verify.
For this problem, I think you need to know the final temperature of the metal? I'm going to assume that its the same final temp as the water, and just set up the basic idea for you.
The main idea is that the energy transfer between the water and the metal should be equal- according to the law of conservation of energy.
deltaE(water) = - deltaE(metal)
(The signs are opposite because here water endothermically absorbs energy and the metal exothermically gives it off)
So now you just need to figure out how to calculate the change in energy for each. That's given by a simple expression using specific heat, mass, and change in transfer:
deltaE = (specific heat of substance)(mass)(change in temperature)
Specific heat is the amount of energy needed to raise 1 gram of a substance by 1 degree celsius. So if a certain plastic has a specific heat of 3 Joules/gram (Every gram of plastic needs 3 joules to raise it by 1 degree C), then to raise 4 grams of it by 2 degrees, each of the 4 grams needs 3 joules per degree, and there are two degrees, so 4x3x2=24 total joules of energy needed.
You equation from earlier now becomes:
(specific heat of water)(mass of water)(temperature change in water) = - (specific heat of metal)(mass of metal)(temp change of metal)
5 of the 6 variables are given to you, and the last one is the specific heat of the metal which you are asked to solve.
The specific heat of water is 4.184 J/g ( if I remember correctly) and while its not in the problem, its something that you should remember because you'll use it quite a bit.
One thing to note is when you calculate the temperature changes of the metal and the water, beware of the proper signs. You know your answer is wrong when your specific heat is negative. Also, if you include your units in your calculations and do dimensional analysis, your units should work out to J/g.
Good luck
the final temp = t
86.3g of water (known heat capacity) is heated from 24.5 to 33.1 deg C
54.3g of metal (unknown heat capacity) is cooled from 99.7 to 33.1 deg C
working in calories-
(33.1-24.5)*86.3*1 = (99.7-33.1)*54.3*X (Heat Capacity of the metal)
8.6*86.3*1 = 66.6*54.3*X
X=0.2052 calories per gram deg
you can convert from calories/g deg to joules/g deg
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