A system dissipates 12 JJ of heat into the surroundings; meanwhile, 28 JJ of work is done on the system. What is the change of t
1 answer:
Given that,
A system dissipates 12 J of heat into the surroundings.
28 J of work is done on the system.
To find,
The internal energy of the system.
Solution,
The first law of thermodynamics is used here. According to this law,
Q is heat and W is work done
Here,
Q = -12 J is the heat dissipated by the system
W = -28 J is the work done on the system
ATQ,
So, the change of internal energy of the system is 16 J.
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1) Physical principles:
a) Total mechanical energy = kinetic energy + potential energy.
b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)
2) Notation:
a) Total mechanical energy: ME.
b) Kinetic energy: KE
c) Gravitational potential energy: PE
∴ ME = KE + PE = constant
3) Solution:
a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.
b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0
c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).
d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.
At that point, KE = 0, and PE = ME.
That is the point of highest altitude and where the speed is zero.
d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.
Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.
The PE may be sketched on the same graph along with the KE and the ME.
The graph is attached. The red line is the ME and the blue line is the PE.
Note that at any point in the graph PE + KE = ME.
