# 29.0 mL of ethanol (density = 0.789 g/mL) initially at 8.3*C is mixed with 37.7 mL of water (density = 1.0 g/mL) initially at 24.6*C in an insulated beaker. Assuming that no heat is lost, what is...

29.0 mL of ethanol (density = 0.789 g/mL) initially at 8.3*C is mixed with 37.7 mL of water (density = 1.0 g/mL) initially at 24.6*C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?

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### 1 Answer

The ethanol will lose heat and the water will gain heat as an equilibrium temperature is established. The heat lost by the ethanol equals the heat gained by the water. The heat transferred by each substance is calculated using

q=mc `Delta` T

where m is the mass, c is the specific heat capacity of the substance and `Delta`T is the change in temperature. Since we don't know the final temperature, we can express the change in temperature of the ethanol as Tf-8.3 degrees C and that of the water as 24.6-Tf.

This gives us

m(ethanol)c(ethanol)(Tf-8.3 degrees) = m(water)c(water)(24.6 degrees-Tf)

The specific heat capacities are:

ethanol: c = 2.44 J/g-degree and water: c = 4.18 J/g-degree

(This is the heat required to raise the temperature of one gram by one degree C)

The masses of water and ethanol are calculated by multiplying volume by density:

mass of ethanol = 29.0 ml x 0.789 g/ml = 22.9 g

mass of water = 37.7 ml x 1.00 g/ml = 37.7 g

(22.9g)(2.44 J/g-deg)(Tf-8.3 deg) = (37.7g)(4.18 J/g-deg)(24.6-Tf)

55.83 Tf - 463.4 = 3880. - 157.7 Tf

(55.83 Tf + 157.7 Tf) = 3880. + 463.4

213.5 Tf = 4343

**Tf = 20.3 degrees C = final temperature**