Question

A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the lanten heats of vaporization of 5 liquids below, determine what the substance is:Acetone: 538,900 J kg-1Ammonia: 1,371,000 J kg-1Propane: 356, 000 J kg-1Methane: 480, 600 J kg-1Ethanol: 841,000 J kg-1A) EthanolB) AmmoniaC) PropaneD) Acetone

Answer 1

Correct answer: C) Propane

Latent heat of vaporization is the amount of heat required to bring about the phase transition from liquid to gaseous state, at its boiling point.

Given mass of the liquid = 28.47 kg

Heat required to boil the liquid =
`10.14*10^(6) J`

The latent heat of vaporization of the liquid =
`(10.14 * 10^(6) J)/(28.47 kg) = 356200 (J)/(kg)`

Latent heat of the liquid is close to that of propane (356000 J/kg).

So, the given liquid is propane.

Answer 2

Answer:- The substance is Propane and right choice is C.

Solution:- 10140000 J of heat is used to boil 28.47 kg at 298 K. Here, the important thing is, the liquid is changing to vapor and the temperature is not changing means it's a phase change. So, the formula used for this is:

`q=m*\Delta H_v_a_p`

where, q is the heat energy, m is mass and
`\Delta H_v_a_p` is the enthalpy of vaporization.

Let's rearrange the formula for
`\Delta H_v_a_p` as:

`\Delta H_v_a_p` =
`(q)/(m)`

Let's plug in the values:

`\Delta H_v_a_p` =
`(10140000J)/(28.47kg)`

`\Delta H_v_a_p` =
`356164(J)/(kg)`

If we round this then it is
`356000(J)/(kg)` . Hence, the right choice is C. propane.