A 3.00-g bullet (c = 0.0305 cal/g . C° = 128 J / kg . C°) moving at 180 m/s enters a bag of sand and stops. By what amount does the temperature of the bullet change if 80% its KE becomes thermal energy that is transferred to the bullet?

Thank ´s fir your help!

35,000

kelvin

To determine the change in temperature of the bullet, we need to use the equation:

ΔQ = mcΔT

where:
ΔQ is the thermal energy transferred to the bullet,
m is the mass of the bullet,
c is the specific heat capacity of the bullet material, and
ΔT is the change in temperature of the bullet.

First, let's calculate the thermal energy transferred to the bullet:

KE_initial = (1/2)mv^2

where:
m is the mass of the bullet and
v is the velocity of the bullet.

Given:
m = 3.00 g = 0.003 kg (converting grams to kilograms)
v = 180 m/s

KE_initial = (1/2)(0.003 kg)(180 m/s)^2
KE_initial ≈ 0.486 J (joules)

Now, we need to find 80% of the initial kinetic energy (KE_initial):

KE_transferred = 0.80(KE_initial)
KE_transferred = 0.80(0.486 J)
KE_transferred ≈ 0.389 J

Now, let's substitute the values into the equation to solve for the change in temperature (ΔT):

ΔQ = mcΔT
ΔT = ΔQ / (mc)

In this case,
m = 0.003 kg (mass of the bullet)
c = 0.0305 cal/g °C * (1 cal/4.186 J) * (1 kg/1000 g) = 0.0305/4186 J/kg °C (converting specific heat capacity from calories to joules and grams to kilograms)

ΔT = 0.389 J / (0.003 kg * 0.0305/4186 J/kg °C)
ΔT ≈ 426.95 °C

Therefore, the temperature change of the bullet is approximately 426.95 °C.