A 90 g steel bullet moves at 230m/s just before inbedding itself wood. If 80% of the initial KE goes into heating the bullet (steel C=460J/kg-C): How hot does the bullet become if it is initially at 27 C? What would the bullet temp be if it was 2X big?
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Anonymous
ANSWERS: 1
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Work out the KE for the bullet eg 45 * 230 ^2. Then multiply that by 0.8 That's the energy that goes into heating it. 0.8 KE = mass * dT * C So dT = 0.8 KE / (Mass * C) mass is 0.09kg and you know C, so work out the temperature yourself. If the bullet was twice as heavy, e.g. 180g, then the dT would be half the value, so you divide thast by two and add it to 27. I expect you'll get an answer in the order of a hundred or so celsius final, but thats a rough (GUESS) off my head
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