A single shot of espresso is produced by passing water at approximately 95 ° C a

Question

A single shot of espresso is produced by passing water at approximately 95 °C and 9 atm through about 7.5 g of ground coffee. A number of oily molecules (including caffeine) are quite soluble in the water under these conditions. A single shot of espresso is approximately 30 mL and contains roughly 50 mg of caffeine. Calculate the net change in energy and entropy for a process that begins with pure water and caffeine at 25°C and ends with 30 mL of espresso at 65 °C (recommended serving temperature).

Potentially useful things that you can know or assume:

i)The molecular weight of caffeine is 194.19 g / mol.

ii)Assume that H is 0 for dissolving caffeine in water and that this is independent of pressure and temperature.

iii)Assume a constant heat capacity of caffeine of 1.2 J / g K. (It actually varies by about 20% from 25 °C to 100 °C, but this is good enough.)

iv) The density of water is T-independent and equal to 1 g / mL.

v) Water is an incompressible fluid (no change in pressure will change the volume of the water).

vi) The heat capacity of water is independent of T, P, and caffeine concentration and is equal to 4.184 J / g K

Explanation / Answer

deltaH = m * c * deltaT

m = mass of water + mass of caffeine

volume of espresso = volume of water + volume of caffeine

30 mL = V + 50 x 10^-3 / 1.23

V = 30 - 0.041 = 29.96 mL

Mass of water = 29.96 * 1 g/mL = 29.96 g

m = 29.96 + 0.05 g = 30.01 g

c = specific heat of caffeine + specific heat of water = 1.2 + 4.184 = 5.384 J/g-K

deltaT = 338 - 298 K = 40 K

deltaH = 30.01 * 5.384 J/g-K * 40 = 6.46 kJ

deltaH = 6.46 kJ

deltaS = S(final) - S(initial)

deltaS = 6.46 x 10^3 / 338 - 0

deltaS = 19.12 J/K

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