Will a mixture of 40 ml water at 100 C is mixed with 80 ml water at 50 C?

Answer 1

Use this formula. q =nCdeltaT

(100 grams)(4.180 J/gC)(Tf - 80 degr.) + (100 g)(4.180 J/gC)(Tf - 40 degr.) = 0

(418Tf - 33440) + (418Tf - 16729) = 0

836Tf = 50169

Temp. final = 60 degrees Celsius

sounds reasonable to me

Answer 2

Regardless of the quantity of water, temperature flows from hot to cold. So 50 grams of water at 60 degrees celcius will drop in temperature and the 100 grams of water at 30 degrees will rise in temperature.

Let the final temperature of the mixture be "T". Heat gained = Heat lost in a closed system (conservation of energy principle)

So (60 - T) x 50 = (T - 30) x 100

or

(60 - T) x 5 = (T -30) x 10

or

(60 - T) = (T -30) x 2

or

60 - T = 2T - 60

or

120 = 3T

Hence T = 40 degrees celcius

Answer 3

(100*30 + 50*60)/(100+50) = (300 + 300)/150 = 40 degrees.

Answer 4

50 x 20 + 75 x 60 = (50+75) T

1000 + 4500 = 125T

T = 5500/125 = 1100/25 = 44 C this only works because both sub. are water and the specific heat cancels out of the problem.

Answer 5

Take the weighted average. That is, multiply each mass by the corresponding temperature, add everything together, then divide the result by the total mass.

Answer 6

The mixing is possible.