Question

Constants A 14.7 g bullet traveling horizontally at 866 m/s passes through a tank containing 14.5 kg of water and emerges with a speed of 544 m/s Part A What is the maximum temperature increase that the water could have as a result of this event? AT 3337.047 C Submit Request Answer

Answer 1

Using energy conservation

Kinetic energy lost by bullet = Internal energy increased due to heat

dKE = Q

dKE = KEi - KEf

dKE = 0.5*m*Vi^2 - 0.5*m*Vf^2

Vi = Initial speed of bullet = 866 m/sec

Vf = final speed of bullet = 544 m/sec

m = mass of bullet = 14.7 gm = 14.7*10^-3 kg

Q = Energy increased due to heat = M*C*dT

M = mass of water = 14.5 kg

C = Specific heat of water = 4186 J/kg-C

So,

dKE = Q

0.5*m*(Vi^2 - Vf^2) = M*C*dT

dT = 0.5*m*(Vi^2 - Vf^2)/(M*C)

Using given values:

dT = 0.5*14.7*10^-3*(866^2 - 544^2)/(14.5*4186)

dT = 0.0549787 C

dT = increase in temperature = 0.0550 C

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