Suppose that 150.7g of ethanol at 22.4°C is mixed with 201.4 g of ethanol at 56.

Question

Suppose that 150.7g of ethanol at 22.4°C is mixed with 201.4 g of ethanol at 56.5°C at a constant atmospheric pressure in athermally insulated vessel. Calculate S andStot for the process.
S at 22.4°C inJ·K-1
answer: 23.31
S at56.5°Cin J·K-1
answer: -22.07
Stot inJ·K-1
answer: 1.24
I'm not sure how to get theseanswers and any help would be greatly apreciated. Suppose that 150.7g of ethanol at 22.4°C is mixed with 201.4 g of ethanol at 56.5°C at a constant atmospheric pressure in athermally insulated vessel. Calculate S andStot for the process.
S at 22.4°C inJ·K-1
answer: 23.31
S at56.5°Cin J·K-1
answer: -22.07
Stot inJ·K-1
answer: 1.24
I'm not sure how to get theseanswers and any help would be greatly apreciated.

Explanation / Answer

                         specificheat of ethanol = 2.46 J/g.oC           let the final temperature of the mixture = T           heat gain by the cold liquid equal to heat lost by the hotliquid.              150.7g * 2.46 J/ g.oC  * (T - 22.4oC) = 201.4 g * 2.46 J/ g .oC* ( 56.5 -T)                               T = 41.84oC              S at 22.4°C = m c logT2/T1                                       = 150.7g *  2.46J/ g .oC * log 41.84oC /22.4°C                                                       S at 56.5°C = 201.4 g * 2.46 J/ g .oC * log 41.84oC / 56.5 oC                       Convert   S in J/oC toJ/K then add the two values to get the total entropychange.

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