A 3.8 kg piece of aluminum at 30 degrees C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20 deg C). Calculate the approximate net change in entropy of the system.

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Aniqa O&#039;Neill · Accepted Answer

We find the final temperature from heat lost = heat gained,  malcal△Tal=mwatcwat△Twat  plug in the values and solve for T  (3.8)(0.22)(30−T)=(1)(1)(T−20)  25.08−0.836T=T−20  45.08=1.836T  T=24.56∘C  Then we find the heat flow from aluminum to water:  Q=malcal△Tal  Q=(3.8)(0.22)(30−24.56)  Q=4.55kcal=4.55×103 cal  Since the heating and cooling do not occur at the same temperatures, we will use the average temperatures to estimate theentropy change.  Twat=Twat+T2=20+24.562=22.28∘C  Tat=Tat+T2=30+24.562=27.28∘C  Now we can calculate the total entropy change  △S=△Sal+△S{wat  △S=(−QTal)+(+QTwat)  , remember we have to convert Celsius to Kelvins,so just add 273.15 to the temperatures  △S=(−4.55×103300.43)+(4.55×103295.43)  △S=0.26calK

A 3.8 kg piece of aluminum at 30 degrees C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20 deg C). Calculate the approximate net change in entropy of the system.

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