Question: The amount of internal energy needed to raise the temperature of .25 kg of water by 0.2 degrees Celsius is 209.3 J. How fast must a 0.25 kg baseball travel in order for its kinetic energy to equal this internal energy?

KE = (.5)(m)(v^2)
209.3 J = (.5)(.25kg)(v^2)
1674.4 = v^2 **square root both sides**
v = 40.92

^Is ANY of that correct? If not, can anyone help? Thanks so much

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  1. I don't see what 1/4 kg of water has to do with a baseball, but you did the problem correctly.

    The specific heat of a baseball is different from that of water, and its mass is 1/7 kg, so what you have calculated in NOT the velocity needed to raise the baseball 0.2 degrees. It is an arbitrary number.

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  2. yes that is correct

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  3. You are right!!

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  4. yep! and its 2017. :)

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  5. yes and its 2019

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  6. Yes, and it’s 2020

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  7. Yep and it’s April 20 Year 2069

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