Question about a bunsen burner lab?

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SUMMARY

The discussion centers on calculating the temperature of a bunsen burner flame by analyzing the heat transfer between heated copper and water. The equation used is -(mass copper)(Cp copper)(T3-T1)=(mass water)(Cp water)(T3-T2), where T1 is the burner temperature, T2 is the initial water temperature, and T3 is the final water temperature. A key point raised is the need to account for the energy required to evaporate water, quantified at 2.26 kilojoules per gram. The suggestion to weigh both the copper and water before and after the experiment is crucial for accurate calculations.

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chemishard93
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Okay, I have a lab write-up due tomorrow and I am completely stumped on one of my conclusion questions.

The lab was an attempt to figure out the temperature of a bunsen burner flame. We did this by using a bunsen burner to heat copper and then immersing the copper into a cup of water and recording the temperature change. The question is asking us to take into account the small mass of water neglected in the equation that was evaporated into hot steam when the copper was immersed.

The equation was: -(mass copper)(Cp copper)(T3-T1)=(mass water)(Cp water)(T3-T2)

T1=burner temperature
T2=intial water temperature
T3=final water temperature

Okay...THE QUESTION I CAN'T FIGURE OUT IS: It takes about 2.26 kilojoules of energy to evaporate each gram of water. Suggest a way to include the evaporated water in your calculation.

I have tried to come up with solutions. However, none of them seem to make much sense...

PLEASE HELP! NEED HELP WITHOUT 3-4 HOURS!
 
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Well... if you were to weigh the copper strip and the water in the calorimeter before heating and immersing the heated strip, you might have a good start on it.
 
We did that, actually. Haha - sorry! Forgot to mention that.
 
And, did you weigh it afterwards?
 

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