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Answer :
The temperature change of the bullet is approximately:
[tex]129.33 \, t{K[/tex]
The correct temperature change of the bullet can be calculated using the formula for kinetic energy and the specific heat capacity of lead.
First, we calculate the initial kinetic energy (KE) of the bullet using the formula:
[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]
where m is the mass of the bullet and v is its velocity. Plugging in the given values:
[tex]\[ KE = \frac{1}{2} \times 0.09 \, \text{kg} \times (182 \, \text{m/s})^2 \][/tex]
[tex]\[ KE = \frac{1}{2} \times 0.09 \times 33124 \][/tex]
[tex]\[ KE = 0.5 \times 0.09 \times 33124 \][/tex]
[tex]\[ KE = 1489.58 \, \text{J} \][/tex]
Since all the kinetic energy is converted to heat, this energy will cause the temperature of the bullet to rise. The temperature change [tex](\( \Delta T \))[/tex] can be found using the formula:
[tex]\[ Q = mc\Delta T \][/tex]
where Q is the heat energy, m is the mass of the bullet, c is the specific heat capacity of lead, and [tex]\( \Delta T \)[/tex] is the temperature change we want to find. Rearranging the formula to solve for[tex]\( \Delta T \):[/tex]
[tex]\[ \Delta T = \frac{Q}{mc} \][/tex]
Substituting the known values:
[tex]\[ \Delta T = \frac{1489.58 \, \text{J}}{0.09 \, \text{kg} \times 128 \, \text{J/kg} \cdot \text{K}} \][/tex]
[tex]\[ \Delta T = \frac{1489.58}{11.52} \][/tex]
[tex]\[ \Delta T = 129.33 \, \text{K} \][/tex]
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Answer: Δθ = 127.4 K
Explanation: by using the law of conservation of energy, the kinetic energy of the bullet equals the heat energy on the plate.
Kinetic energy of bullet = mv²/2
Heat energy = mcΔθ
Where m = mass of bullet = 0.09kg, v = velocity of bullet = 182 m/s, c = specific heat capacity of lead bullet = 130 j/kgk
Δθ = change in temperature
mv²/2 = mcΔθ
With 'm' on both sides of the equation, they cancel out each other, hence we have that
v²/2 = cΔθ
v² = 2cΔθ
Δθ= v²/2c
Δθ = (182)²/2×130
Δθ = 33124/260
Δθ = 127.4 K
