Question

If you combine 360.0mL of water at 25.00C and 110.0mL of water at 95.00C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water. Answer in Celcius.

Answer 1

Assume final temperature 'x ^oC '

The energy amount going out of the warm water is equal to the energy amount going into the cool water.

So, q_lost = q_gain

We know that,   q = (mass) (Δt) (C_p)

As density of water is 1gm/cc so volume of water=mass of water

Now,    (mass) (Δt) (C_p) = (mass) (Δt) (C_p)

Where q_lost on the left side and q_gain on the right side.

So, by substitution, we then have:

(110.0) (95.0 - x)(4.184) = (360.0) (x - 25.0) (4.184)

or, (110.0) (95.0 - x) = (360.0) (x - 25.0)

or, 10450 -110.0x = 360.0x- 9000

or, 470x = 19450

or, x= 41.4 ^oC

Thus the final temperature of the mixture= 41.4 ^oC