If the final temperature is 23.5C what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water is 1.00g/mL)
2 Answers

heat gained by cold water = heat lost by hot water
(m Cp dT) cold = (m Cp dT) hot
cancel Cp since Cp cold wate r= Cp hot water
(m dT) cold = (m dT) hot
mass cold water = mass hot water x dT hot water / dT cold water
mass cold water = (21.4 mL x 1.00 g/ml) x (35.0 – 23.5) / (23.5 – 18.2) = 46.4 g cold water
= 46.4 g x (1 mL / 1.00 g) = 46.4 mL
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if you recognized that density = mass x volume —–> m = ρV…
then …
m cold = m hot x dT hot / dT cold
ρV cold = ρV hot x dT hot / dT cold
assume ρ cold water = ρ hot water and cancel ρ
V cold = Vhot x dT hot / dT cold…
ie V cold = 21.4 mL x (35.0 – 23.5) / (23.5 – 18.2) = 46.4 mL

The unknown ΔT (Temp. difference) = 23.5 – 18.2 = 5.3°C.
The known ΔT = 35 – 23.5 = 11.5°C.
(21.4mL x 1mL/g = 21.4g)
Unknown x 5.3°C = Known 21.4g x 11.5°C.
Unknown = (21.4 x 11.5) / 5.3 = 46.4g. (46.4mL)
(Check: 46.4g x 5.3°C = 21.4g x 11.5….i.e. 246 = 246).