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 $\frac{J}{kg}$

Ammonia: 1,371,000 $\frac{J}{kg}$

Propane: 356,000 $\frac{J}{kg}$

Methane: 480,600 $\frac{J}{kg}$

Ethanol: 841,000 $\frac{J}{kg}$

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 =$\frac{10.14 * 10^{6} J}{28.47 kg} = 3,56,000 \frac{J}{kg}$

So the substance boiled is propane.