Question

A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the latest heats of vaporization, what substance is this?

Answer 1

A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the Latent heats of vaporization of some of the substances:

Acetone: 538,900
`(J)/(kg)`

Ammonia: 1,371,000
`(J)/(kg)`

Propane: 356,000
`(J)/(kg)`

Methane: 480,600
`(J)/(kg)`

Ethanol: 841,000
`(J)/(kg)`

Calculating the latent heat of vaporization of the given substance:

Heat required =
`10.14 * 10^(6) J`

Mass of the substance boiled = 28.47 kg

Latent heat of vaporization =
`(10.14 * 10^(6) J)/(28.47 kg) = 3,56,000 (J)/(kg)`

So the substance boiled is propane.