Question

A 50.0-g piece of iron at 152 C is dropped into 20.0 g H20(I) at 90 C in an open, thermally insulated container How much water would you expect to vaporize, assuming no water splashes out? The specific hoats of

Answer 1

The final temperature of iron and water must be equal to 100⁰C otherwise no vaporization can take place.

Assuming no heat lost to the environment:

Heat lost by iron = M*C*dT = 50*0.45*(152-100) = 1170 J

Heat gained by water = M*C*dT = 20*4.21*(100-90) = 842 J

At this point, excess heat energy = 1170-842 = 328 J

This much energy is used in vaporizing the water.

Assume that 'x' g of water is vaporized.

Using equation:

m*dH_vap = Excess energy

Putting values:

x*(40700/18) = 328

Solving we get:

x = 0.145 g

Hope this helps !