let observer A be in frame S and observer B in frame S'.(adsbygoogle = window.adsbygoogle || []).push({});

let the watch of A be synchronized with the watch of B.

let S' move at a constant speed of 0.913c relative to S.

let t0 = the time that B reads by his own watch and t = the time that A reads by the watch of B.

if t0 = 1s, then t = 1s/sqrt(1 - 0.913^2) = 2.45s

this means that A observes that 1s by the watch of B is made up or consists of 2.45s by his own watch (the watch of A). in other words, A observes that it takes 2.45s by his own watch (the watch of A) to make 1s by the watch of B. in other words A reads both his own watch and the watch of B and concludes that his own watch is 2.45 faster than the watch of B.

let's now say that the following thermodynamic process takes place in S (S is at rest relative to A):

a system in its initial state consists of two paper cups (C and F, separated by a barrier that insulates one cup from the other) inside a box made up of walls that insulate the inside of the box from the outside. in the initial state of the system C contains 0.57kg of hot water (TiC = 363 k) and F contains 0.57kg of cold water (TiF = 283 k). when A removes the barrier between the two cups, he noticed that eventually TfC = TfF = 323 k. A also noticed that the net entropy change of the system from initial to final state is +37J/K. the specific heat of water is c = 4,190 J / kg * K, and the heat capacity of the paper cups is insignificant.

if this thermodynamic process takes place in S', what would be the observation of A (S' moves at a speed 0.913c relative to A)? what would A say about the net entropy change of the system?

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# Question concerning thermodynamics and special relativity

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