Scienc Posted October 21 Posted October 21 Is this resolution wrong? Since the value of Wad is -4.68 KJ, applying this value to the expression ln(pi/pf) should yield a positive result of 0.750 instead of -0.750, correct? This is due to the minus sign in the equation.
studiot Posted October 21 Posted October 21 (edited) Surely delta T is negative ? Edited October 21 by studiot
Scienc Posted October 21 Author Posted October 21 47 minutes ago, studiot said: Surely delta T is negative ? Isn't it? Since the final temperature is 300 K and the initial temperature is 450 K, we can calculate the change in temperature (ΔT) as follows: ΔT = (Tf - To) = (300 - 450) = -150 K. By the way, I think the person who solved this problem included a negative sign in front of the first equation (ΔU = nCpΔT) because they used the temperatures in the reverse order in their calculations.
studiot Posted October 22 Posted October 22 Delta T = Final temperature - Initial temperature = 300 - 450 = -150 which is negative as I said. But you seem to have used it the other way round (450 - 300) when you substitute numbers into the formula.
Scienc Posted October 22 Author Posted October 22 3 hours ago, studiot said: Delta T = Final temperature - Initial temperature = 300 - 450 = -150 which is negative as I said. But you seem to have used it the other way round (450 - 300) when you substitute numbers into the formula. I believe the individual who solved this problem included a negative sign in front of the first equation (ΔU = nCpΔT) because they reversed the order of the temperatures in their calculations. The equation itself does not contain a negative sign.
sethoflagos Posted October 23 Posted October 23 (edited) 23 hours ago, Scienc said: I believe the individual who solved this problem included a negative sign in front of the first equation (ΔU = nCpΔT) because they reversed the order of the temperatures in their calculations. The equation itself does not contain a negative sign. Engineers and chemists generally have opposing views on whether thermodynamic work performed BY a system ON its surroundings is a positive or negative quantity. Engineers tend to the historic thermodynamics tradition of treating it as positive. Quoting from https://en.wikipedia.org/wiki/Work_(thermodynamics) Quote Sign convention [edit] In the early history of thermodynamics, a positive amount of work done by the system on the surroundings leads to energy being lost from the system. This historical sign convention has been used in many physics textbooks and is used in the present article.[25] According to the first law of thermodynamics for a closed system, any net change in the internal energy U must be fully accounted for, in terms of heat Q entering the system and work W done by the system:[14] ΔU=Q−W. [26] An alternate sign convention is to consider the work performed on the system by its surroundings as positive. This leads to a change in sign of the work, so that ΔU=Q+W. This convention has historically been used in chemistry, and has been adopted by most physics textbooks.[25][27][28][29] It's a persistent source of confusion. Negative signs often appear and disappear in these calculations as the various equations are modified to fit a preferred convention. Edited October 23 by sethoflagos sp
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