# When 155 mL of water at 26 degrees C is mixed with 75 mL of water at 85 degrees C, what is the final temp?

Could someone SHOW me the steps to working this one out.

Assume no heat is lost to the surroundings, d of water is 1 g/mol

• 155ml(1g/ml) = 155g H2O

75ml(1g/ml) = 75g H2O

(mc∆t) = - [(mc∆t)]

(155g)(4.184J/g•℃)(tf - 26℃) = - [(75g)(4.184J/g•℃)(tf - 85℃)

648.52tf - 16861.52 = -[313.8tf - 26673]

962.32tf = 43534.52

tf = 45℃

Q1 = Q2

m1Cp(Tf-T1) = m2Cp(T2-Tf)

m1(Tf-T1) = m2(T2-Tf) Cp cancels on both sides

where

m1 and m2 are the masses of the different waters, so you will need the density of water to determine this: d = m/V. You have both volumes:

V1 = 155 mL

V2 = 75 mL

T1 = 26 C

T2 = 85 C

Solve for Tf = final temperature.

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