A%25252525252520bottle%25252525252520with%25252525252520volume%252525252525200.2

Question

A%25252525252520bottle%25252525252520with%25252525252520volume%252525252525200.25%25252525252520m%2525252525255E3%25252525252520is%25252525252520initially%25252525252520evacuated%25252525252520and%25252525252520closed.%25252525252520Eventually%2525252525252C%25252525252520the%25252525252520cap%25252525252520of%25252525252520the%25252525252520bottle%25252525252520is%25252525252520removed%2525252525252C%25252525252520and%25252525252520the%25252525252520atmospheric%25252525252520outside%25252525252520air%25252525252520rushes%25252525252520into%25252525252520the%25252525252520bottle.%25252525252520After%25252525252520some%25252525252520time%2525252525252C%25252525252520the%25252525252520bottle%2525252525252C%25252525252520which%25252525252520remains%25252525252520open%2525252525252C%25252525252520will%25252525252520have%25252525252520air%25252525252520inside%25252525252520in%25252525252520equilibrium%25252525252520with%25252525252520the%25252525252520outside%25252525252520air%25252525252520(same%25252525252520temperature%25252525252520and%25252525252520pressure).%25252525252520Determine%25252525252520the%25252525252520heat%25252525252520that%25252525252520crosses%25252525252520the%25252525252520surface%25252525252520of%25252525252520the%25252525252520bottle%25252525252520during%25252525252520the%25252525252520process%2525252525252C%25252525252520if%25252525252520any%2525252525252C%25252525252520in%25252525252520KJ.%25252525252520The%25252525252520outside%25252525252520air%25252525252520is%25252525252520at%2525252525252027%25252525252520degrees%25252525252520Celsius%25252525252520and%25252525252520100%25252525252520kPa

Explanation / Answer

dU=?Q?VdP

where dU=cp*dt = change in internal energy

dT= 300 k

total energy = internal energy change + V (dP)

assuming no air

internal energy =Cp * dT

dq=0.25 * 100000Pa + 300 kj

total= 25 kj +300 kj =325 kj

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