posted by mike .
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
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.
yes that is correct
You are right!!
yep! and its 2017. :)