A 4.00-g lead bullet is traveling at a speed of 200 m/s when it embeds in a wood post. If we assume that half of the resultant heat energy generated remains with the bullet, what is the increase in temperature of the embedded bullet? (specific heat of lead = 0.0305 kcal/kg⋅°C, 1 kcal = 4 186 J)

 To calculate the increase in temperature of the lead bullet, we first need to determine the kinetic energy (KE) that the bullet had before embedding in the wood post. We can use the kinetic energy formula:

KE = 1/2 mv^2

where m is the mass of the bullet (converted to kilograms for consistency with the specific heat unit) and v is its velocity.

m = 4.00 g = 0.004 kg (since there are 1000 grams in a kilogram) v = 200 m/s

KE = 1/2 * 0.004 kg * (200 m/s)^2 = 0.002 kg * 40,000 (m^2/s^2) = 80 J

Now, we assume that half of this energy is converted to heat, which stays with the bullet. So the heat energy (Q) is:

Q = KE / 2 = 80 J / 2 = 40 J

Next, we have to convert the specific heat of lead from kcal/kg°C to J/kg°C:

Specific heat (c) = 0.0305 kcal/kg⋅°C * 4186 J/kcal ≈ 127.673 J/kg⋅°C

Now, we can use the formula for heat transfer to calculate the temperature change (ΔT):

Q = m * c * ΔT

Solving for ΔT:

ΔT = Q / (m * c) = 40 J / (0.004 kg * 127.673 J/kg⋅°C) = 40 J / (0.510692 J/°C) ≈ 78.33°C

So, the increase in temperature of the embedded bullet is approximately 78.33 degrees Celsius.