A coffee-cup calorimeter contains 50.0g ofwater at 60.51 degree C. A 12.4 g piec

Question

A coffee-cup calorimeter contains 50.0g ofwater at 60.51 degree C. A 12.4 g piece of graphite at 24.21 degreeC is placed in the calorimeter. The final temperature of the waterand the carbon is 59.02 degree C. Calculate the specific heat ofcarbon. The specific heat of water is 4.18J/g degree C.

The answer should be .721 J/g degree C..

I need some help on getting to this answer and the steps.

I tried to find the heat for water but I don't know if that wouldbe part of solving the question

• 13 hours ago
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A coffee-cup calorimeter contains 50.0g ofwater at 60.51 degree C. A 12.4 g piece of graphite at 24.21 degreeC is placed in the calorimeter. The final temperature of the waterand the carbon is 59.02 degree C. Calculate the specific heat ofcarbon. The specific heat of water is 4.18J/g degree C.

The answer should be .721 J/g degree C..

I need some help on getting to this answer and the steps.

I tried to find the heat for water but I don't know if that wouldbe part of solving the question

Explanation / Answer

So for this problem, you need to know a couple ofthings. First you need to know the equations to find heat inthermochemistry problems. The first equation is: q= mass x specific heat x tempertatue (q=MCT). Now for this problem, you are going to set up the equationslike so: M 1C 1 T1 =M2C2 T 2 Now before you start the problem, you need to specify what islosing heat and what is gaining heat. For a Calorimeter problem,you can make either equation negative. For this problem lets makethe second equation negative ( remember you could make the firstequation negative. It doesn't really matter) M 1C 1 T1 = -(M2C2 T 2) Now lets plug in the values: (50g H20)(4.18J/g)(59.02-60.51) = - (12.41gCarbon)(X)(59.02-24.21) (-311.41) = -(431.99x)     now solve forx: -311.41 = -431.99x x = 0.721 J/g Hope this helped. Don't forget to give Karma please. Now before you start the problem, you need to specify what islosing heat and what is gaining heat. For a Calorimeter problem,you can make either equation negative. For this problem lets makethe second equation negative ( remember you could make the firstequation negative. It doesn't really matter) M 1C 1 T1 = -(M2C2 T 2) Now lets plug in the values: (50g H20)(4.18J/g)(59.02-60.51) = - (12.41gCarbon)(X)(59.02-24.21) (-311.41) = -(431.99x)     now solve forx: -311.41 = -431.99x x = 0.721 J/g Hope this helped. Don't forget to give Karma please.

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