If you have 50gr. of H2O at 90degrees and 20gr. of H2O at 10degrees, what will be the change in temperature?

Question

Please help and give an answer as soon as possible?!?!

please help me this is for a project im doing in chemistry and i need to know how to do it.

Answer 1

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Suppose when 50 gram hot water at 90 degree is mixed with 20 gram cold water at 10 degree, the final temperature is ' X '

Specific heat of water = c calorie /gram /degree Celsius

Hot water gives heat when its temperature falls from 90 to X degree.

Cold water takes heat when its temperature rises from 10 to X degree. From principle of calorimetry,

Heat lost by hot water =heat gained by cold water

50*c*[90 -X] =20*c*[X - 10]

450 - 5X = 2X - 20

7X=470

X=470/7=67.14 degree Celsius

Temperature of hot water decreases by 160 / 7 =22.86 degree Celsius

Temperature of cold water increases by 400 / 7=57.14 degree Celsius
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Answer 2

Let's try a new approach without specific heats etc...: -

Mixture = 70 grams
ΔT = 80°C
50g at 90°C = 72% of mixture = 72% of 80°C
= 57.6°C will be gained by the cold water.
= 10 + 57.6 = 67.6°C

20g at 10°C = 28% of mixture = 28% of 80°C
= 22.4°C will be lost by the hot water.
= 90 - 22.4 = 67.6°C
Heat gained = Heat lost.
Final temp. = 67.6°C

(I don't think we need be concerned with about ½°C ??? )
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Answer 3

50(90 - T) = 20(T - 10)
4500 - 50T = 20T - 200
70T = 4700
T = 67.14°C
90 - T = 22.86°C
T - 20 = 57.14°C