Energy needed to boil mercury vs. water

[Aprilshowers]

I need help with a problem I have struggled with for several days.
Here is the problem:
The specific heat of mercury is .03 cal/g degrees C, and it's boiling point is
357 degrees C. The specific heat of water is 1 cal/g degrees C. It takes
65 calories of energy to vaporize one gram of mercury and 540 calories to
vaporize 1 gram of water. If both substances begin a room temperature
(about 22 degrees C), Does it take more energy to boil a gram of mercury or
a gram of water?
I found where the boiling point of water is 100 degrees C...but there is so
much information here that I cannot determine what is needed or not, and
what order to process the information...can anyone advise?

 

[Nylex]

You need to use [itex]Q = mc(T_{f} - T_{i})[/tex] for both the water and the mercury. $T_{f}, T_{i}$ are the final and initial temperatures (which you are given for both water and mercury), c is the specific heat capacity and m is the mass (if you use J K^-1 kg^-1 and kg for those quantities, you'll get an answer in J).[/itex]

 

[OlderDan]

You need to use [itex]Q = mc(T_{f} - T_{i})[/tex] for both the water and the mercury. $T_{f}, T_{i}$ are the final and initial temperatures (which you are given for both water and mercury), c is the specific heat capacity and m is the mass (if you use J K^-1 kg^-1 and kg for those quantities, you'll get an answer in J).[/itex]

$<br /> You also need the latent heat of vaporization given for each sustance. The question is asking the heat needed to raise the temperature of the liquid <b>and</b> to vaporize it.$

 

[Nylex]

Ahh ok, yeah.