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 [tex] \frac{J}{kg} [/tex]
Ammonia: 1,371,000 [tex] \frac{J}{kg} [/tex]
Propane: 356,000 [tex] \frac{J}{kg} [/tex]
Methane: 480,600 [tex] \frac{J}{kg} [/tex]
Ethanol: 841,000 [tex] \frac{J}{kg} [/tex]
Calculating the latent heat of vaporization of the given substance:
Heat required = [tex] 10.14 * 10^{6} J [/tex]
Mass of the substance boiled = 28.47 kg
Latent heat of vaporization =[tex] \frac{10.14 * 10^{6} J}{28.47 kg} = 3,56,000 \frac{J}{kg} [/tex]
So the substance boiled is propane.
