A 2-kilogram object falls 3 meters. a. How much potential energy did the object have before it fell? b. How much work was accomp
1 answer:
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Complete question:
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.
(a) Find the available kinetic energy of the bullet. J
(b) Find the heat required to melt the bullet. J
Answer:
Part (a) the available kinetic energy of the bullet is 323.32 J
Part (b) the heat required to melt the bullet is 216.17 J
Explanation:
Given;
mass of the bullet = 3.53 g = 0.00353 kg
velocity of the bullet = 428 m/s
initial temperature of the bullet = 40.0°C
final temperature of the bullet = 327°C
specific heat capacity, c= 128 J/(kg · °C)
latent heat of fusion, Hf = 24.5 kJ/kg
Part (a) the available kinetic energy of the bullet. J
KE = ¹/₂ × mv²
KE = ¹/₂ × 0.00353 × 428²
= 323.32 J
Part (b) the heat required to melt the bullet. J
This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.
Quantity of heat required to raise the temperature of the bullet:
Q = mcΔT
= 0.00353 × 128 × (327-40)
= 0.00353 × 128 × 287
= 129.68 J
Quantity of heat required to melt the bullet:
Q = 0.00353 × 24500
= 86.49 J
TOTAL energy required to melt the bullet = 129.68 J + 86.49 J
= 216.17 J