Thermal Physics: Dry Steam Heat 200g Water from 25°C to 95°C

[DJ-Smiles]

Homework Statement

Dry steam is used to make a cup of coffee by bubbling it through water. If the
steam is at 100°C, what mass of steam must be used to heat 200 g of water from
25°C to 95°C?

Homework Equations

Not quite sure but I think:
Q= mCΔT
Qcold=-Qhot

The Attempt at a Solution

Ok so I started out by trying to use Qcold=-Qhot, and this was what I did:

Hot:
m=??
C=2020 (this is what the textbook said the specific heat of steam was)
Ti=100°C
Tf= 95°C (because I assumed that they would end up the same temp because of equilibrium)

Cold:
m=0.2kg
C=4200 (textbook said this was specific heatr for water)
Ti=25°C
Tf=95°C

So then I subbed in values to come up with:

4200x0.2(95-25)=-(2020m(95-100))
58800=10100m
m=5.82kg

this is a ridiculous number and the textbook says that the answer is 26g.

Please help me understand this.

 

[gneill]

Think "phase change"

 

[Chestermiller]

You left out the heat of condensation of the steam.

 

[DJ-Smiles]

yeah thanks guys I realized just then the answer should have been 58800=2270100m, m= 58800/2270100= 0.0259kg=25.9g=26g. Thanks for that guys I usually do really well in physics so when i can't understand something I start to stress ahah. Much love and God Bless