Answer:
The change in temperature of the lead bullet is 333.06 ⁰C
Explanation:
Given;
mass of the lead bullet , m = 3.5 g = 0.0035 kg
velocity of the bullet, v = 292 m/s
specific heat capacity of the lead bullet, c = 128 J/kg.◦ C
let Δθ = change is temperature
Kinetic energy of the bullet is calculated using the formula below;
K.E = ¹/₂mv²
K.E = ¹/₂(0.0035)(292)²
K.E = 149.212 J
If all the kinetic energy of the lead bullet is converted to thermal energy and none leaves the bullet, then the temperature change is calculated using the formula below;
K.E = Q
where;
Q is the quantity of heat generated by the moving bullet or say heat capacity of the lead bullet
Q = mcΔθ
where;
m is mass of the bullet
c is specific heat capacity of the lead bullet
Δθ is the change in temperature
Δθ [tex]= \frac{Q}{mc} = \frac{149.212}{0.0035*128} = 333.06 \ ^oC[/tex]
Therefore, the change in temperature of the lead bullet is 333.06 ⁰C
Answer:
333.06 °C
Explanation:
From the question, and applying the law of conservation of energy
Kinetic energy of the lead bullet = Thermal energy of the lead
1/2mv² = cmΔt.................... Equation 1
Where m = mass of the lead bullet, v = kinetic energy of the lead bullet, c = specific heat capacity of the lead bullet, Δt = change in temperature of the lead bullet.
make Δt the subject of the equation
Δt = 1/2mv²/(cm)
Δt = 1/2v²/c...................... Equation 2
Given: v = 292 m/s, c = 128 J/kg.°C
Δt = 1/2(292²)/128
Δt = 333.06 °C
Hence the change in temperature of the lead bullet = 333.06 °C
