# An unknown volume of water at 18.2 C is added to 21.4mL of water at 35C. ?

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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)

• 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).

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