# Homework Help: Thermal physics need help.

1. May 17, 2012

### DJ-Smiles

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
Not quite sure but I think:
Q= mCΔT
Qcold=-Qhot

3. 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 text book says that the answer is 26g.

2. May 17, 2012

### Staff: Mentor

Think "phase change"

3. May 17, 2012

### Staff: Mentor

You left out the heat of condensation of the steam.

4. May 17, 2012

### DJ-Smiles

yeah thanks guys I realised 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