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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 5 liquids below, determine what the substance is. acetone: 538,900 j kg-1 ammonia: 1,371,000 j kg-1 propane: 356,000 j kg-1 methane: 480,600 j kg-1 ethanol: 841,000 j kg-1

Sagot :

Correct option is C

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.

The substance is Propane :

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× Δ[tex]H_{vap}[/tex]

where, q is the heat energy, m is mass and  Δ[tex]H_{vapour}[/tex]  is the enthalpy of vaporization.

Let's rearrange the formula for  as:

Given mass of the liquid = 28.47kg

Heat required to boil the liquid = 10.14 ×[tex]10^{2} J[/tex]

Let's plug in the values: [tex]\frac{10.14*10^{2}J }{28.47Kg} =356164\frac{J}{Kg}[/tex]

Rounding off 356164 will turn into 356000 J[tex]Kg^{-1}[/tex]

Hence, the right choice is C. propane 356000 j[tex]Kg^{-1}[/tex]

What is latent heat ?

Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

Learn more about latent heat of vaporization :

brainly.com/question/24203857

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