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what amount of water at 40 degree celsius is to be mixed with 250 grams of ice at 10 - 10 degree Celsius so that equally premium temperature is 25 degree Celsius
Asked by imunilu786 | 29 Oct, 2023, 02:35: PM
Given that,
Mass of ice, mi = 250 g
Temperature of ice, Ti = -10°C
Inital Temperature of water, Tw=40°C
Final temerature, Tf =25°C

Value of constants:
Latent heat of fustion, Lf = 334 J/g
Specific heat capacity of water, cw= 4.18 J/g°C
Specific heat capacity of ice, ci= 2.09 J/g°C

Now,
Heat lost by water =  Heat gained by Ice
mcw(Tw- Tf) = 250×ci× (0 - Ti) + 250×Lf + 250×cw(Tw- Tf)
By substituting the values we get:

m = (250×2.09×10 +250×334 + 250×4.18×25)/(4.18×15)
∴ m ≈ 1832 g
Answered by Jayesh Sah | 07 Nov, 2023, 09:49: AM

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