A lead ball, with an initial temperature of 25.0oC, is released from a height of 125.0 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature (in degrees C) of the ball after it hits. Data: clead = 128 J/kgoC.
Please I need the answer and the equation used to do this thanks!
h t tp s: / /studyres. com/doc/877 2671/physics-homework-%231-kinematics-di splacement- a ndamp %3B-velocity
I tried Hope this helps!
To find the temperature of the ball after it hits, we need to use the principle of conservation of energy. The potential energy of the ball at the initial height is converted into thermal energy (heat) when it hits the surface.
The potential energy of an object is given by:
PE = mgh
Where:
PE is the potential energy
m is the mass of the ball
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height from which the ball is dropped
In this case, the potential energy is converted into thermal energy. We can calculate the thermal energy using the equation:
Q = mcΔT
Where:
Q is the thermal energy
m is the mass of the ball
c is the specific heat capacity of the lead (given as 128 J/kgoC)
ΔT is the change in temperature
Since all the energy of the fall goes into heating the lead, we can equate the two equations and solve for ΔT:
mgh = mcΔT
Canceling the mass (m) on both sides of the equation:
gh = cΔT
Now, let's plug in the values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 125.0 m (initial height)
c = 128 J/kgoC (specific heat capacity of lead)
Substituting the values into the equation:
(9.8 m/s^2) * (125.0 m) = (128 J/kgoC) * ΔT
Calculating:
1225 J = (128 J/kgoC) * ΔT
Finally, let's solve for ΔT:
ΔT = (1225 J) / (128 J/kgoC)
ΔT ≈ 9.57 oC
Therefore, the temperature of the lead ball after it hits the surface would be approximately 9.57oC.