Physics Help?

Question

A 15-g lead bullet traveling at 220 m/s passes through a thin iron wall and emerges at a speed of 160 m/s. If the bullet absorbs 50 percent of heat generated,

(a) What will be the temperature rise of the bullet?
(b) Of the ambient temperature is 20 C, will any of the bullet melt, and if so, how much?

Answer 1

The kinetic energy lost is
1/2 m v1^2 - 1/2 m v2^2 =
1/2*0.015*220^2 - 1/2*0.015*160^2 = 363-192 = 171 J

The lead will absorb 171 J /2 = 85.5 J

The heat capacity of Lead is 26.650 J/(mol·K)
15 g of lead is equal to 15/82 mol = 0.183 mol

So if 26.650 J will increase the temperature of 1 mole by 1 degree
85.5 J will increase the temperature of 0.183 mole by

85.5/0.183/26.650 = 17.5 degrees Kelvin

So the bullet temperature will reach 20 + 17.5 = 37.5 degrees
It will be luke warm at body temperature and will certainly not melt.

Answer 2

First, calculate the change in kinetic energy based on the change in velocity. The problem does no really say how the energy loss occurs, but we will just assume that it is all heat loss. So, half the number you calculated above is the energy that goes to heating the bullet, use the bullets specific heat and caluclate the temperature change. Add that to 20 and compare to the melting point of lead for the second part.

Answer 3

TE = KE + Q = kinetic energy plus heat (Joules).

Before impact: TE = KE = 1/2 mv^2; where m = .015 kg, v = 220 m/sec.

After impact (conservation of energy): TE = KE + Q; so that Q = TE - KE = 1/2 mv^2 - 1/2 mu^2 = 1/2 m(v^2 - u^2); where u = 160 m/sec. Solve for Q, which is the total heat generated in Joules (Newton-meters).

a) Heat absorbed by bullet q = 1/2 Q; so q = 1/2 Q = k(T1 - T0) and k = the specific heat under constant pressure for lead and T1 - T0 = delT = the required temperature rise. Thus, delT = 1/2Q/k. (You need to look up k = cp for lead.)

b) With T0 = 20 deg C; the raised temperature is T1 = delT + T0. You need to look up the melting point for lead to answer this one. (I presume there are no phase changes and consequent heat losses other than the solid to liquid phase change.)

NOTE: T is usually in Kelvin, but C can be used here because we are dealing with changes in temperature and del K = del C. If you need K, just convert using 0 deg K = -273 deg C as your basis for conversion.

Answer 4

Kinetic energy lost = 0.5 m ( V^2 - U^2)

= 171J.

Bullet absorbs 85.5 J.= ms x rise in temperature.

(15/1000) x 126 x rise in temperature = 85.5

rise in temperature = 45.23 degree centigrade.

Melting point of lead is 340K.

Temperature of bullet = 293 + 45.23 = 338.23.
Hence it will not melt.

Answer 5

Sorry - i dont remember my formulae