A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
1 answer:
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
You might be interested in
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
