Brightstorm Q & A : A 3.0 g bullet (c=.0305 cal/g and Degree=128 Degree C) moving at 180 m/s enters a bag of sand and stops. By what amount does the temp of bullet change if all its KE becomes heat energy added to bullet

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Find the kinetic energy 1/2mv^2 Equate to mc(change in temperature)=1/2mv^2 c *Change in tem=v^2/2 solve for change in tem and find the tem

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chitra003 April 13, 2011

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KE = 1/2 mv^2 = 48.6 Jq = mC(delta T) 48.6 J = 3 g * .127612 J/g*deg * (delta T) delta T = +127 degrees C

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Seth_Warncke June 07, 2011

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