In an experiment to determine the specific latent heat of water, steam of 100°C was passed into water Contained in a well lagged copper calorimeter....

      

In an experiment to determine the specific latent heat of water, steam of 100⁰C was passed into water
Contained in a well lagged copper calorimeter. The following measurements were made:
-Mass of calorimeter =40g
-Mass of water + calorimeter = 135g.
-Final mass of calorimeter + water +condensed steam =148g
-Final temperature of the mixture =52⁰C.
Given that specific heat capacity of water = 4200Jkg^-1 k^-1
Specific heat capacity of copper = 390Jkg^-1 k^-1.
Determine the:
i)Mass of condensed steam
ii)The heat gained by the calorimeter and water if the initial temperature of the calorimeter and water
is 24⁰C
iii)Given that LV is the specific latent heat of vaporization of steam, write an expression for the heat
given out by steam.
iv)Determine the value of LV in (iii) above.
v)State the assumption made in the experiment.

  

Answers


Davis
(i)Mass of condensed steam:
=(148-135)g
=13g

ii)Heat gained by water = mc?= 0.095 x 4200(52-24)=11172J?
Heat gained by calorimeter =mc?=0.06 x 390 x 28 =655.2J
Total heat gained =11172 + 655.2
=11827.2 Joules

iii)Heat given out by steam=Heat lost by condensing steam + heat lost by condensed steam
=0.013 x LV + 0.013 X 4200 X (100-52)
=0.13LV + 2620.8

iv)Total heat lost = Total heat gained
0.013LV + 2620.8 =11827.2
0.013LV =11827.2-2620.8
LV=9206.4/0.013
=708,184.6154Jkg-1

v)No heat loses



Githiari answered the question on September 16, 2017 at 08:18


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