Calculating Latent Heat of Steam: Mass, Temp & Thermal Capacity

[Salerk]

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

 

[Astronuc]

Note the age/date of the OP - Feb 22, 2008.

Firstly, one should use units with values.

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

It's difficult to understand what one might have done to get an answer of 11275 (units?).

The answer should be about 2260 J/g, or 2.26 MJ/kg or 540 cal/g for the latent heat of steam (at 1 atm).

Now interestingly, I've seen values for the conversion between kcal and J ranging from 4183 J to 4187 J. The 'official' value has changed slightly over the decades. Doing a conversion with an online calculator/convertor with Google, I find 1 kcal = 4184 J, and referring to CRC's Handbook of Chemistry and Physics, 68th Ed., 1987-1988, one finds 1 cal = 4.1868 J. This could be related to improvements/differences in measurement technique.

So one reason to do basic physics experiments is to become familiar with procedures and measurement techniques, learning about sources of uncertainty/error, and learning how to process the raw data into something meaningful, and learning how measurement technique might affect the results.

Some relevant material - https://www.sciencedirect.com/topics/engineering/latent-heat-of-steam

Note in one citation, "As the latent heat of steam at atmospheric pressure is 2257 kJ/kg . . ." as compared to 2260 kJ/kg. The difference of 4 kJ/kg in 2260 kJ/kg represents an uncertainty/error of 0.00177 or 0.177%.

 

[PeroK]

More homework being handed in late!