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A copper calorimeter of mass 60g is filed with 100g of water at 250C. Steam at a normal temperature and pressure ( N.T.P) is passed thought...

A copper calorimeter of mass 60g is filed with 100g of water at 250C. Steam at a normal
temperature and pressure ( N.T.P) is passed thought the water until a temperature 450C is attained.
The final mass of calorimeter and the contents was found to be 163.5g. Calculate the specific latent
heat of vaporization ‘l’ of water (Specific heat capacity for water is 4200JKg^-1 and for copper is 378Kg^-1K^1)

Answers

Davis
Heat gained by calorimeter
= 60 x 10-3 x 378 ( 45 – 25) J;
= 453.6 J
Heat gained by water
= 100 x 10-3 x 4.200 ( 45 – 25J;
= 8.400J
Heat lost by condensing steam = m/
( 163.5 – 160 ) x 10^-3/J
= 3.5 x 10^-3 x / J
Heat lost 3.5 g of ( condensed steam) water cooling to 45^0C
3.5 x 10^-3 ( 100 – 45) x 4,200;
= 808.5J
Heat given = heat gained
Hence:
3.5 / x 10^-3 + 808.5 J = 453 6J + 8,400J;
= 2.3 x 10^-6 J/Kg;
Githiari answered the question on September 19, 2017 at 07:09

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