Calculating Latent Heat of Water Vaporization

[series111]

Homework Statement

from the following observations made during an experiment, determine the specific latent heat of vaporisation of water.

initial mass of water 400g, final mass of water 450g (the extra is condensed steam)

mass of calorimeter 200g, initial temp of water & calorimeter 16 C

final temp of water & calorimeter 70 C , specific heat capacity of water 4200j/kg C

specific heat capacity of calorimeter 420 j/kg C

HINT heat lost by steam= heat gained by water + heat gained by calorimeter

Homework Equations

quantity of heat = mass x specific heat capacity x temperature change

The Attempt at a Solution

let suffix c = calorimeter
w = water

mc cc tc = mw cw tw + fmw cw t

-200 x 420 ( 70 tc) = 400 x 4200 ( 16 t) + 450 x 420 ( 16 t)

- 84000 ( 70 tc) = 168,0000 ( 16 t) + 189,000 (16 t)

-84000 ( 70 tc) = 1869000 ( 16 t)

- 84000 c + 5.88 x 10 6 = 1869000 c 29.904 x 10 6

5.88 x 10 6 +29.904 x 10 6 = 1869000 c 84000 c


35.784 x 10 6 = c (1869000 + 84000 ) = 1953000 c


c = 35.784 x 10 6 / 1953000 = 18.322 C


MY TUTOR HAS SAID MY FORMULA IS INCOMPLETE BUT I DONT NO WHERE I HAVE GONE WRONG PLEASE HELP MARK.

 

[series111]

can anyone help ?

 

[Moetasim]

Dear Mark,

Thank you for sharing your attempt at solving the problem. Your formula is indeed incomplete, as it does not take into account the heat lost by the steam. Here is a revised formula that takes into account all the heat transfers:

Heat gained by water = Heat lost by steam + Heat gained by calorimeter

(mc + mw) (cw) (tf - ti) = ms (Lv) + mc (cc) (tf - ti)

Where:
mc = mass of calorimeter
mw = mass of water
cw = specific heat capacity of water
tf = final temperature (in this case, 70°C)
ti = initial temperature (in this case, 16°C)
ms = mass of steam
Lv = specific latent heat of vaporization of water
cc = specific heat capacity of calorimeter

Rearranging this equation, we get:
Lv = [(mc + mw) (cw) (tf - ti) - mc (cc) (tf - ti)] / ms

Now, let's plug in the values from the problem:
mc = 200g
mw = 400g
cw = 4200 J/kg°C
tf = 70°C
ti = 16°C
ms = 50g (calculated from the difference between initial and final mass of water)
cc = 420 J/kg°C

Substituting these values into our formula, we get:
Lv = [(200 + 400) (4200) (70 - 16) - 200 (420) (70 - 16)] / 50
Lv = 2,940,000 J/kg = 2.94 x 10^6 J/kg

Therefore, the specific latent heat of vaporization of water is 2.94 x 10^6 J/kg.

I hope this helps clarify the incomplete formula in your attempt. Keep up the good work in your scientific endeavors!

Best,