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Introductory Physics Homework Help
Not sure if I calculated this correctly (Find the final temperature)
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[QUOTE="PhyIsOhSoHard, post: 4693420, member: 408052"] Oh yes, you're right. Then I get: [itex]T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600-2256000\cdot 2}{2\cdot 4190+2\cdot 910}=-319 °C[/itex] That's clearly not correct though so I'm assuming this isn't the correct method that I'm using. How do you know the final temperature is 100 °C? Is it because it can't get any higher than 100? If we want to calculate how much water liquid remains we need to figure out how much x amount is vaporized [itex]xm_{water}L_{water}[/itex] at 100 degrees: [itex]m_{Al}c_{Al}(100-T_{Al})+xm_{water}L_{water}+m_{water}c_{water}(100-T_{water})=0[/itex] Solve for x: [itex]x=\frac{-m_{Al}c_{Al}(100-T_{Al})-m_{water}c_{water}(100-T_{water})}{m_{water}L_{water}}[/itex] Insert values: [itex]x=\frac{-2\cdot 910\cdot (100-600)-2\cdot 4190\cdot (100-20)}{2\cdot 2256000}=0.053[/itex] That means a total of [itex]0.053\cdot 2kg=0.106kg[/itex] water is vaporized. So the water that remains at 100 degrees is: [itex]2kg-0.106kg=1.894kg[/itex] But what now? [/QUOTE]
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Introductory Physics Homework Help
Not sure if I calculated this correctly (Find the final temperature)
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