If 44.25 kJ of heat are released by changing 150. g of liquid mercury to mercury vapor,
what is the latent heat of vaporization for mercury in kJ/kg?
29,500 kJ/kg
0.3 kJ/kg
295 kJ/kg
3.4 kJ/kg
I believe the answer is 295 but I am not sure
44.25 kJ /0.150 kg= you have it.
To find the latent heat of vaporization for mercury, you need to divide the heat released (44.25 kJ) by the mass of the substance that underwent the phase change (150 g).
The formula for latent heat is:
Latent Heat of Vaporization = Heat Released / Mass
In this case:
Latent Heat of Vaporization = 44.25 kJ / 150 g
However, to get the answer in kJ/kg, you need to convert the mass of mercury from grams to kilograms. One kilogram is equal to 1000 grams.
Mass (in kg) = Mass (in g) / 1000
So, the final formula becomes:
Latent Heat of Vaporization = 44.25 kJ / (150 g / 1000)
Simplifying this calculation, you have:
Latent Heat of Vaporization = 44.25 kJ / 0.15 kg
Now, performing the division:
Latent Heat of Vaporization = 295 kJ/kg
Therefore, your initial assumption is correct. The latent heat of vaporization for mercury is indeed 295 kJ/kg.