Question

Solve for the final temperature of the mixture of 400.0 mL of water at 25.00 °C and add 140.0 mL of water at 95.00 °C. Use 1.00 g/mL as the density of water.

Answer 1

(mass)(ΔT)(Cs) = (Cs)(ΔT)(mass)
(140.0g)(95.00−x)(4.184)=(4.184)(x−25.00)(400.0g)
Solve for x
55647.2 - 585.76x = 1673.6x – 41840
-585.76x – 1673.6x = -41840 – 55647.2
-2,259.36x = -97,487.2
-2,259.36x / -2,259.36 = -97,487.2 / -2,259.36
X = 43.15