# Chemistry

To begin the experiment, 1.65g of methane CH is burned in a bomb calorimeter
containing 1000 grams of water. The initial temperature of water is 18.98°C. The
specific heat of water is 4.184 J/g °C. The heat capacity of the calorimeter is 615
J/ °C. After the reaction the final temperature of the water is 36.38°C.
2. Calculate the change in temperature, AT. Show your work.
How would I even start this off?
Thanks

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1. Note the methane is CH4 and not CH.
I'm a little confused about what you want. If you want the change in temperature of the water, it is 36.38 - 18.98 = ?
But there must be more to the problem and finding delta T for water.

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DrBob222
2. So would I use the 17.4 to find the heat absorbed by water? Im confused

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3. yes. I am assuming you are trying to find the heat of combustion of CH4.
CH4 + O2 ==> CO2 + H2O + heat.
So the water absorbs heat and the calorimeter absorbs heat. The heat absorbed by the water is
q = mass H2O x specific heat H2O x delta T.

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DrBob222
4. How would I find the heat absorbed by methane? qcal = Ccal • ΔT

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5. heat is not absorbed the the methane. However, to find the heat absorbed by the calorimeter you have it correct; i.e., Ccal*delta T. Then total heat absorbed is q from H2O + q from calorimeter. The total will be the heat generated by the combustion of 1.65 g CH4. Convert that to heat absorbed by a mole of CH4 (16 grams that is) instead of the 1.65 g. I believe you're getting there.

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DrBob222
6. thank you so much

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7. Can you respond with the calculation of the change in temp ? Please

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