A lump of lead with mass 0.50 kg is dropped
from a height of 20 m onto a hard surface. It
does not rebound but remains there at rest for
a long period of time.
What are:
a) ∆Q,(b) ∆Wand (c) ∆Ufor the lead during
this process?
d) What is the temperature change in the lead
immediately after the impact? [specific
heat capacity of lead = 128 kJ/kg K]
What is heat mass0.5kg from dropped height20m.remains at rest
Alump of lead
Physics
Alump of lead with mass 0.50kg is dropped from a height of 20m on to a hard surface
a lump of lead with mass 0.50kg is dropped from a height of 20m onto a hard surface. it does not rebound but remains there at rest for a long period of time
what are change of Q,W,U for the lead during this process
What is the answer
q=98j , w=0andT=586k
To determine the values of ∆Q, ∆W, and ∆U for the lead during this process, we can apply the principles of thermodynamics and work-energy theorem.
a) ∆Q represents the change in thermal energy. In this case, since the lead lump is at rest for a long period of time after impact, there is no transfer of heat into or out of the lump. Therefore, ∆Q is zero.
b) ∆W represents the work done on the lead. When the lead lump is dropped and hits the hard surface, it comes to a stop. This means that the kinetic energy of the lead is converted into work done by an external force to bring it to rest. The work done on the lead can be calculated using the formula: ∆W = mgh, where m is the mass of the lead (0.50 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (20 m) from which it was dropped. Plugging in the values, we have: ∆W = (0.50 kg)(9.8 m/s²)(20 m) = 98 J.
c) ∆U represents the change in internal energy. Since the lead comes to rest and remains there for a long period of time, there is no change in internal energy. Thus, ∆U is also zero.
d) The temperature change in the lead can be determined using the formula: ∆Q = mC∆T, where ∆Q is the change in thermal energy (which is zero), m is the mass of the lead (0.50 kg), C is the specific heat capacity of lead (128 kJ/kg K), and ∆T is the change in temperature we want to find. Rearranging the formula to solve for ∆T, we have: ∆T = ∆Q / (mC). Since ∆Q is zero, the temperature change ∆T is also zero. Therefore, there is no temperature change in the lead immediately after the impact.
In summary:
a) ∆Q = 0
b) ∆W = 98 J
c) ∆U = 0
d) ∆T = 0 (no temperature change)