# Calorimetry homework help

1. Aug 3, 2010

### mizzy

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

A copper container of mass 0.080kg and specific heat of 387Jkg-1K-1 contains 0.30kg of water and 0.040kg of ice at 0degrees. Steam at 100 degrees is passed into the water and its temperature stabilizes at 20.0degrees. Find the mass of the water left in the container.

2. Relevant equations
Q = mc(T2 - T1)

Q = mL

3. The attempt at a solution
I don't know where to start. Can someone help me in how to approach this question?

2. Aug 3, 2010

### arkofnoah

Re: calorimetry...LOST!!!

The heat from the steam is used to melt the ice and to raise the temperature of water (including that from the melted ice) to 20.9 degrees. The question is what is the mass of steam needed to supply this amount of heat? Find that out (don't forget to take into account the latent heat of vaporisation) and add up the total mass of the water (the melted ice, the condensed steam and the original water).

3. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

I'm not getting it. Do you have to find Q for each phase?? Like find Q from ice to water and then find Q from water to steam??

Can someone explain this to me?

THANKS

4. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

okay think of the heat lost:

1) condensation of steam into water
2) decreasing the temperature of the 100 degrees water to 20 degrees.

the heat gained:

1) melting of ice into water
2) increasing the temperature of 0.340 kg of water from 0 degrees to 20 degrees.

heat lost = heat gained

5. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

hmmmmmmmmmm...

the copper container loses heat, right? so using the equation: Q = mc(Tf-Ti)

Q = 0.080kg * 387J/kgK (293K - 273K)
= 619.2J

is that a start??

6. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

7. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

now what?

i don't know what to do with the water. there are two things happening, steam warms up the water and melts the ice. is that correct??

8. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

yes that's correct, but you'd have to leave anything to do with steam to the end because you don't know how much steam condensed (you are supposed to work that out eventually). keep trying i'll just prompt you along the way :D

9. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

would this be my equation to find the mass of water left?

Heat lost = heat gained

Qcopper + Qsteam = Qice + Qwater
mc(tf -tf) + mc(tf - ti) = mc(tf - ti) + mc(tf - ti)

10. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

nope. have you learnt about the latent heat of vaporization and fusion?

this is more like it:
Qcopper + Qsteam + latent heat of vaporisation = Qice + Qwater + latent heat of fusion

11. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

yes, that's the energy needed to change a phase.

so in the equation above, latent heat of vaporisation is the phase change from gas to liquid? and then latent heat of fusion is the change from ice to liquid.

is that right??

ok...let me do the work first and i'll post what i did. THANKS!!

12. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

oh wait wait!!

Qsteam + latent heat of vaporisation = Qcopper + Qice + Qwater + latent heat of fusion

the copper should be under heat gained since it's at 0 degrees initially.

and the Qsteam actually refers to the amount of heat lost when the condensed water drops to 20 degrees, from 100 degrees. similarly with Qice.

13. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

k...don't know if i'm doing this right. I thought I knew what i was doing...hehe =(

Let me explain the left hand side of the equation first:

HEAT LOST:

Qcopper + Qsteam + latent heat of vaporisation

mc(Tf - Ti) + mc(Tf - Ti) + mLv

where the following is given: mass of copper, specific heat of copper, final temperature 20degrees and initial is 0degrees (are my temperatures right???) mass of steam is unknown, specific heat of steam is known, final temp = 20 and inital temp = 100. not too sure of the phase change. its Q = mL, is mass the mass of steam???

14. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

the copper container gained heat, not lost heat. it was initially at thermal equilibrium with the water and ice.

15. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

Ok.

so then heat lost:

Qsteam + latent heat of vaporisation

mc(Tf - Ti) + mL

Tf = 20degrees, Ti = 100degrees, L and c is known and m is what we are looking for, RIGHT??

16. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

yup. carry on for the other side of the equation.

17. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

Ok. So heat gained:

Qcopper + Qice + latent heat of fusion + Qwater

mc(Tf - Ti) + mc(Tf - Ti) + mLf + mc(Tf - Ti)

where mass of copper, ice and water are given.
for copper: Tf = 20, Ti = 0
for ice: Tf = 100, Ti = 0
for water: Tf = 20, Ti = 0

the mass in the equation, mLf, m is mass of ice...right??

are my temperatures correct??

18. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

for ice: Tf = 100, Ti = 0

ALL final temperature should be 20. The zeroth law of thermodynamics.

19. Aug 5, 2010

### mizzy

Re: calorimetry...LOST!!!

k.

I jsut want to clarify what the question is asking for. it's asking for the mass of water left in the container. What we solved for is mass of steam. In the second posting you said to add up the mass of ice, steam and water. why??

20. Aug 5, 2010

### arkofnoah

Re: calorimetry...LOST!!!

because... the ice melted and the steam condensed. that means the amount of water in the container increased, right? we just add up the mass of the ice, steam and water to get the total mass of water left in the container. how else can you interpret this question?