A lead bullet with a mass of 8.00 g and traveling at 5.00 times 10^4 cm/s is imb

Question

A lead bullet with a mass of 8.00 g and traveling at 5.00 times 10^4 cm/s is imbedded in a wooden block weighing 1.000 kg. If both the block and the bullet were initially at 25.0 degree C, what is the final temperature of the block containing the bullet? The specific heat of wood is 0.50 cal/g degree C; that of lead is 0.030 cal/g degree C. 1 cal = 4.184 J, and J = kg m^2/s^2. { Assume no heat loss to the surroundings. You may also assume that the heat absorbed by the bullet is negligible! Why?}

Explanation / Answer

Given:

Mass of bullet = 8.00 g = 0.008 kg

Speed = 5.00 E4 cm/ s = (5.00 E4 cm/s) x (1m/ 100 cm) =5.00 E2 m/s

Initial temperature of the wood and bullete = 25 deg C

Lets calculate internal energy

Internal energy = kinetic energy of bullet

Kinetic energy of bullet = ½ mv2

=(1/2 ) 0.008 kg x (5.00 E2 m/s)2

=2400 kg m2 s2 = 2400 J

Internal energy of wood = 2400 J

Internal energy of wood = specific heat of wood x (Tf- Ti)

2400 J x 1 cal / 4.184 J = (0.50 cal / g deg C ) x (Tf- 25 deg C)

(Tf- 25 deg C) = 1147.2

Tf = 1147.2 + 25 deg C

= 1172.22 deg C

Final temperature = 1172.22 deg C

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