A drinking glass 0.02 kg contains 200gms of water at 20^0 C. A mass of 0.04 kg of ice at 0^0 C is dropped into the...

      

A drinking glass 0.02 kg contains 200gms of water at 20^0C. A mass of 0.04 kg of ice at 0^0 C is
dropped into the glass. Determine the final temperature of the mixture. Specific heat capacity of
glass = 670kg^-1K^-1. (Give your answer to correct one decimal place)

  

Answers


Davis
Let final temperature be Q2
Heat lost by water = 4200 x 0.2 ( 20- Q2 )
Heat lost by glass = 0.2 x 670 x (20 - Q2 )
Heat gained by ice = 0.04 x 334 x 103
Heat gained water = 0.04 x 4200 ( Q2 – 0)
Heat lost = Heat gained.
4200 x 0.2 (20 - Q2 ) + 0.2 x 670 x ( 20 - Q2 ) = 0.04 x 334 x 103 + 0.04
X 4200 ( q2 – 0 )
Q2 = 5.36^o C
Githiari answered the question on September 19, 2017 at 11:34


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