Please Answer it now...................
10g of ice at 0 °C is added to 20g of water at 15 °C.What is the equilibrium temp.of mixture?
(Given specific heat capacity of water = 1cal/g°C and specific latent heat of fusion of ice is= 80cal/g)
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Heat lost by water = Heat gained by ice
m s (T2-T) = mL + ms(T-T1)
20 (15-T) = 10(80) + 10(1/2) T
300-20T = 800+5T
-500 = 25T
T = -20 °C
RajivShastry:
heat gained by ice = ms(T-T1)
How it is 20(15)/5+20
By equating this and solving, we get above eqn
oops I'm sorry I totally forgot about latent heat
I will edit my answer
Thats was the mistake.....
Please do it soon .I need it today then exlpain it
I have updated the answer.
How there is 1/2
specific heat of ice = 1/2
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