# Specific heat (I'm not sure if it's capacity or latent)

1. Jan 15, 2012

### gabloammar

1. The problem statement, all variables and given/known data
Part a. A 500 W kettle contains 300g of water at 20°C. Calculate the time it would take to raise the temperature of the water to boiling point.

Part b. The kettle is allowed to boil for 2 minutes. Calculate the mass of water that would remain in the kettle.

State any assumptions you make.

(Specific heat capacity of water = 4.18x103 J kg-1 °C-1, specific latent heat of vaporisation of water = 2.26x106 J kg-1.)

2. The attempt at a solution
I have the answer for part a. It's (rounded) about 201 seconds.

Problem: I don't know where to start from for part b. Someone help.

2. Jan 15, 2012

### gabloammar

Oh and I'm making two assumptions.
1. There's no energy lost to the surroundings.
2. All the vapour formed escapes the kettle.

3. Jan 15, 2012

### technician

I agree with your time for part (a)
Once the water is boiling there is no increase in temperature ( boils at 100C) so all of the energy supplied ( in 2 minutes) is used to convert liquid water into water vapor ( steam)
The usual assumptions about no heat loss etc.

4. Jan 15, 2012

### gabloammar

Yep yep I can think about that. Fine. But, the only formula I've got in my head is E=mL, and the only way I'm thinking of finding the energy change [E in the equation] is power equals work done by time taken. Using that, power being 500 W and time taken being 120 seconds, the work equals 60kJ [I'm not even sure if I should be doing that or not].

So I've got energy and I've got specific latent heat. What next? Put 'em in the formula?

mass equals energy by specific latent heat, and I get 37.7 [is this kg or grams?].

Seriously, I'm COMPLETELY at sea with this.

5. Jan 15, 2012

### gabloammar

Oh btw the answer is supposed to be 273g, rounded off to 270g. No idea where it comes from though.

6. Jan 15, 2012

### technician

You are doing the right thing.
60kJ in 2 mins is correct so the mass of steam = ( 60 x 10^3)/2.26 x 10^6
which is 0.027 kg
(where did you get the answer to be 273g?)

Last edited: Jan 15, 2012
7. Jan 15, 2012

### gabloammar

I could've sworn I was doing the same thing over and over again and not getting that value! Or maybe I was looking at the energy in a wrong way [I have the value 59892.5 J written on the notebook, no clue where I got it from though]. But thanks! Question solved :D

8. Jan 15, 2012

### technician

great....don't you wish all problems were so easy

9. Jan 15, 2012

Of course :p