Question

95.2 Kg of a liquid absorbs 4.8 x 107J of heat a

as it boils. What is its heat of vaporization (in joules/gram)?

Answer 1

mass = 95.2 kg (since all of the liquid was vaporized)
heat absorbed = 4.8 x 10^7 J
mass = heat absorbed / heat of vaporization

Solving for the heat of vaporization:

heat of vaporization = heat absorbed / mass

heat of vaporization = (4.8 x 10^7 J) / (95.2 kg)

heat of vaporization ≈ 504,201 J/kg

Therefore, the heat of vaporization of the liquid in question is approximately 504,201 J/kg, or 504.2 J/g.

Answer 2

Explanation:

1000g=1kg

xg=95.2kg

x=95200g

Heat of vapourisation=the thermal energy required for vaporization divided by the mass of the substance that is vaporizing.

mass of the substance=95200g

thermal energy=4.8* 107J

Heat of vapourisation= 4.8*10^7j / 95200g = 504.201680672

To 2 decimal places= 504.20J/g