Calculating Heat Transfer in Mercury and Water: Homework Problem and Solution

[Nimbalo]

Homework Statement

One day Susan collected the mercury from the thermometers around her house and found she had 3.50 cubic centimeters of mercury (density= 13.6 g/cm^3) at room temperature of 28C. She places the mercury in a small bag and drops it into liquid nitrogen (-196C).a. Calculate how much heat energy would be required to change this frozen block of mercury into vaporized mercury at 500C. The specific heat of solid mercury is 134, the specific heat of liquid mercury is 139, and the specific heat of gaseous mercury is 104.

b. What mass of liquid nitrogen was boiled away when she lowered the mercury's temperature to -196C?

Homework Equations

(delta)Q=mc(delta)T

The Attempt at a Solution

I did part a. I got as an answer 45137.6J of energy. However, I'm not quite sure how to set up part b. What equation should I use?EDIT: this seems to be related to the next question:

A construction worker on a rooftop 20m high drops a hot .150kg iron rivet at 500C into a bucket containing 6kg of water at 20C. Assuming no heat is lost to the surroundings or the bucket, and that no water splashes out, what is the final temperature of the rivet and the water?

I'm just not really sure which equation to use...

 

[Delphi51]

You will need to know the amount of heat needed to change a kg of liquid nitrogen into nitrogen vapour. This would be called the "heat of vaporization" of nitrogen and should be available in Wikipedia.

In the first part, you would also need the heat of vaporization of mercury. And the heat needed to liquefy solid mercury (heat of fusion).