In an experiment to determine the specific latent heat of vaporization of water, steam at 1000c was passed into water contained in a well-lagged copper...

      

In an experiment to determine the specific latent heat of vaporization of water, steam at 1000c was passed into water contained in a well-lagged copper calorimeter. The following measurements were made:
Mass of calorimeter = 50g
Initial mass of water = 70g
Final mass of calorimeter + water + condensed steam = 123g
Final temperature of mixture = 300C

(Specific heat capacity of water = 4200 J kg -1K and specific heat capacity for copper = 390 J kg -1 K-1)
Determine the:
i) Mass of condensed steam
ii) Heat gained by the calorimeter and water
iii) Given that L is the specific latent heat of evaporation of steam
I. Write an expression for the heat given out by steam
II. Determine the value of L.

  

Answers


Wilfred
(i)
(I) 123 – 120 = 3g or 0.003 kg

(II)
Heat for water
0.070 x 4,200 x 25
= 7,350J

Heat calorimeter
0.05 x 390 x 25
= 487.5J

Total = 487.5 + 7,350 = 7,837.5J

(ii) I. ML + MC? ? = 7837.5
II. 0.003L + 0.003 x 4200 x 70 = 7837.5
0.003L + 882 = 7837.5
L= 2.3185 x 106

Or L= 2.32 x 106J/kg
Wilfykil answered the question on March 26, 2019 at 14:10


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