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?

1/2 m v^2 * .8=mcDeltaTemp

DeltaTemp=.4*180^2/128

notice mass of the bullet divides out, so no change of units is needed for specific heat.

To find 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, and
ΔT is the change in temperature of the bullet.

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

KE = (1/2)mv^2

where:
KE is the kinetic energy of the bullet,
m is the mass of the bullet, and
v is the velocity of the bullet.

Given:
mass of the bullet (m) = 3 g = 0.003 kg
velocity of the bullet (v) = 180 m/s

KE = (1/2)(0.003 kg)(180 m/s)^2
KE = 0.486 J

Since 80% of the kinetic energy becomes thermal energy, we can calculate the thermal energy transferred to the bullet:

ΔQ = (0.8)(0.486 J)
ΔQ = 0.3888 J

Now, let's use the equation ΔQ = mcΔT to find the change in temperature:

0.3888 J = (0.003 kg)(128 J/kg°C)(ΔT)
ΔT = 0.3888 J / (0.003 kg)(128 J/kg°C)
ΔT = 107.6389°C

Therefore, the temperature of the bullet changes by approximately 107.64°C.

To determine the change in temperature of the bullet, we can use the equation Q = mcΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the initial kinetic energy (KE) of the bullet. KE is given by the equation KE = 0.5 * m * v^2, where m is the mass and v is the velocity.

Given:
Mass of the bullet (m) = 3.00 g = 0.003 kg
Velocity of the bullet (v) = 180 m/s

Now let's calculate the initial kinetic energy:
KE = 0.5 * 0.003 kg * (180 m/s)^2

Next, we need to determine the thermal energy (Q) transferred to the bullet. We are given that 80% of the initial kinetic energy becomes thermal energy, so we can multiply the initial kinetic energy by 0.80 to find the thermal energy.

Thermal energy transferred (Q) = 0.80 * Initial kinetic energy

Now we need to find the change in temperature (ΔT). Rearranging the equation Q = mcΔT, we have:
ΔT = Q / (mc)

Given:
Specific heat capacity (c) = 0.0305 cal/g.°C = 128 J/kg.°C
Mass of the bullet (m) = 0.003 kg
Thermal energy transferred (Q) = 0.80 * Initial kinetic energy

Now we can calculate the change in temperature:
ΔT = (0.80 * Initial kinetic energy) / (0.003 kg * 128 J/kg.°C)

This calculation will give us the change in temperature of the bullet.