- #1

Salerk

- 9

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## Homework Statement

Ive been tasked with working out the Latent heat of steam

the following are all my results and values used.

Mass of the calorimeter and stirrer: 0.068Kg

Thermal Capacity of the Calorimeter and Stirrer : 25.772 JK-1

Mass of water and Calorimeter and stirrer: 0.145Kg

Mass of water = 0.077

Thermal capacity of water: 323.4

Thermal capacity of calorimeter and water: 349.172

Starting Temperature: 19.5 Degrees C

Ending temperature 56 Degrees C

Mass of calorimeter & water after passing steam 0.151Kg

Mass of the condensed steam 0.006Kg

Heat gained by the water: 12744.778

Latent heat capacity of steam: 11635.978 Jkg

## Homework Equations

The equations i had to use are

Thermal Cap of Calorimeter and stirrer = weight x 379 (I can't read my tutors hand wrighting, but i think that's the thermal capacity of copper or along them lines, just know the value 379 to be correct, least I hope it to be)

The Thermal Capacity of water I have as (Total weight - Weight before water added) * 4200

For heat gained by water we where given the equation:

(Thermal Capacity of Water + Thermal Capacity of the Calorimeter) * temprature rise (((From now on referd to as Ct)))

The final equation (and the one I think might be wrong) is this for the calculation of the Latent heat capacity of steam

Ct(T2-T1) - Ms Cw ( 100-T2) / Ms

Ct ((see above))

T2 56 Deg C

T1 19.5 Deg C

Ms = Mass of condenced steam (0.006 kg)

and Cw I've got down as 4186, I think this is ment to stand for the Thermal Capacity of Water, esp if Ct stands for the CT of water + Calorimeter. If I am right then it should be 318.136 not 4186

## The Attempt at a Solution

Well most of it I've done above. but i make the answer to be

343.9(56-19.5)-0.006*4186(100-56) / 0.006

geting me the answer of 11275.

Ive been told that its ment to be 2256000 by my tutor but another said it should be 2256. There are a few 000 missing between the, but neather knew the units (yea, I know) so i looked around and I foud that its being said to be 2260J per Gram

This is where I am stuck.

Ive just got no idear if I've done this wrong, if I've been told wrong values, so trying to deside if the result I've got is right or wrong is sending me crazy

If anyone can clear this up for me that would be fantastic.

and the final question would be. what's the equation to work out Molar latent heat capacity of steam.?

Thanks very much

Ritchie