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A lead bullet weighing 18.0g and travelling at 500 m/s is embedded in a wooden block of 1.00kg. If both the bullet and the block were initially at 25.0 oC, what is the final temperature of block containing bullet? Assume no temperature loss to the surroundings.

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1 Answer

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Solution : –

We have, kinetic energy of bullet is converted into heat  = 1/2 mv2

=  1/2 ×18×10–3×(500)2

=  2.25×103 J

=  (2.25×103)/(4.184×103) kcal

=  0.538 kcal

Also, mS = mS for bullet + mS for wood

= 18×10–3×0.030 + 1×0.50

And  q = K.E.= mS∆T

Therefore,  ∆T  =  K.E./(mS)

=  0.538/(18×10–3×0.030 + 1×0.50)

=  1.08 K

=  1.08° C

Therefore, final temperature  = 25.0° C + 10.8° C

=  26.08° C

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