Heating curve Q

[040111333]

Could someone please help me on this problem?

Q - Heating a 3.5 gram sample of unknown liquid at atmospheric pressure generates the heating curve below.

What is the most likely identity of the compound?

a) Ethanol (Cp, gas = 2.3 kJ/kg·K; ∆Hvap = 840 kJ/kg)
b) Water (Cp, gas = 1.9 kJ/kg·K; ∆Hvap = 2260 kJ/kg)
c) Ammonia (Cp, gas = 2.2 kJ/kg·K; ∆Hvap = 1370 kJ/kg)
d) Propane (Cp, gas = 1.7 kJ/kg·K; ∆Hvap = 356 kJ/kg)

The answer is C) but I have no idea why...

This is part of what it says in the explanation and I didn't understand this:
Given we have 3.5 grams of sample and vaporization required ~4,800 J, the heat of vaporization for our unknown is 4,800 J/3.5 g = 1,370 J/g (ammonia is the unknown and our solution). How did they get 4800 J and what formula is used to get delta H of ammonia?

Can anyone help me?

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[amy_k]

They got 4,800J for vaporization from the horizontal portion of the graph at -33 degrees C where the heat is increased from 116J to 4910J (4910J - 116J = 4794J), but the temperature of the sample remains the same. If you remember at phase changes we get a horizontal line on a heating curve because the heat is used for the phase transition rather than increasing the temperature. So we have 4800J used for vaporization and we want to get the heat of vaporization in J/g so we simply divide by the mass of the sample: 4800J/3.5g = 1,370J/g = 1,370kJ/kg which is ammonia.