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Imagine a bus travelling at a speed of 20m/s towards the positive x-axis with respect to ground frame. That bus is quite long(just for a safer Side) and two persons 'A'and 'B' located inside that bus and thus part of a single frame(not moving relative to each other). And there is a third person 'C'(take C as ground frame for instance) standing outside.'A' has a fundamental particlein its hand(which he is going to throw). At time 't=0', 'A' throws fundamental particle towards negative x-axis(opp. of bus's motion) with a velocity of 5m/s with respect to 'A' and 'B'. Now 'A' and 'B'both will say that the work done on the particle is positive(as from their perspective the velocity is increased) but 'C' will say that the work done by 'A'on that particle is negative as velocity has been decreased( initial velocity of ball=20m/s but final velocity is= 20-5= 15m/s towards the positive x-axis). So positive work done means providing(giving) energy to an object and negative work done means taking energy from the particle. "If giving or taking energy affects the temperature of an object", how will the temperature of that particular ball will change with time? (this is very important for me please don't close this question)

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There are two different kinds of energy here: kinetic energy and thermal energy. And you also have to be careful not to mix up frames of reference. Sure I can pick an arbitrary frame, but I have to be consistent and I cannot switch between them without acceleration which would introduce additional energy transfers.

From C's perspective, A and B both have non-zero kinetic energy $\frac{1}{2}mv^2$ but C is at rest and has zero kinetic energy. But from A and B's perspective, the opposite is true: they do not have kinetic energy but C does. However, they will all agree on each other's temperatures. If an object has kinetic (or potential) energy, it does not change their temperature.

However, let us suppose A and B each have clay ball and they throw them in unison, each at 5 m/s, so they collide and come to rest (at least within the frame of A & B). The two clay balls will experience an inelastic collision and stick together. This will convert kinetic energy to thermal energy and the mass will warm up. Now, let's say the balls were 2 kg each to make the math nice. According to $\frac{1}{2}mv^2$, each clay ball has $\frac{1}{2}*2*5^2$ or 25 Joules of kinetic energy. After colliding, the mass has 50 additional Joules of thermal energy it did not have before and it warms up.

Let's switch to C's frame. A throws a clay ball at 15 m/s and B throws theirs at 25 m/s. So A's ball has $\frac{1}{2}*2*15^2$ or 225 Joules, and B's ball has $\frac{1}{2}*2*25^2$ or 625 Joules. When they strike each other, they combine to have a total of 850 Joules. But it's now a 4 kg clay mass traveling at 20 m/s. So $\frac{1}{2}*4*20^2$ is 800 Joules of kinetic energy, leaving 50 Joules to be converted into thermal energy.

Regardless of frame, everyone agrees that the clay collision converted 50 Joules of kinetic energy into thermal energy. For this reason, when I'm doing calculations, I tend to pick a frame of reference that makes the math easier. A and B's frame is simpler to calculate in and I still get the same answers.

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  • $\begingroup$ I think you (or possibly me) wrong. Temperature is the average total(internal) energy of each molecule. Hence, the increase in kinetic energy will indeed (must) affect temperature... please correct me if I am wrong. $\endgroup$
    – Swayam Jha
    Jan 25 at 14:54
  • $\begingroup$ And please justify your statement that regardless of frame the thermal energy transfer would be constant. Give mathematical proof if possible. $\endgroup$
    – Swayam Jha
    Jan 25 at 14:57
  • $\begingroup$ Apologizes, I'm not sure if I can satisfy your request. If you were one of my students, I'd want to have a long talk at a whiteboard to go over various concepts. I'm not sure I can provide a mathematical proof other than reworking my example above with variables but it would be just as valid. And if I recall, internal energy, by definition, excludes kinetic energy (and potential energy from external forces like gravity or magnetism.) Beyond those clarifications, I'm not going to be much use. Sorry. $\endgroup$ Jan 25 at 15:30
  • $\begingroup$ So sir internal kinetic or internal potential energy do have impact on potential? $\endgroup$
    – Swayam Jha
    Jan 25 at 15:59

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