Question

Thermodynamics: You have a (mostly) insulated carafe in which you store your coffee. Coffee can be modeled as saturated water for this problem. The internal volume of your carafe is 1 L and you seal the lid when it contains 900 g of coffee (mostly liquid, some vapor) at 90 ◦C.

1. What is the mass [g] of water vapor in the carafe?

You are very busy at work and forget to drink your coffee. By the end of the day, it has cooled to 35 ◦C.

2. How much energy [kJ] was lost through the walls of the carafe?

3. What is the pressure [kPa] of your now barely warm coffee?

Answer 1

Solution :-

Given - V = 1L = 0.001 m³

T = 90+273 =363 K

As standard we know that, P = 92.5 mm Hg (for standard liquid)

1mm Hg = 133.32 pa

so that P = 12332.32 pa

and For Standard gas, R = 8.3144 J mole^-1 K^-1

By Ideal Gas equation we have-

P V = n R T

n = 12332.32 x 0.001/(8.314 x 363)

n = 0.04053 mol

since the molar mass of standard liquid,M = 18.01528g mol^-1

mass of vapour, m = M x n = 0.0453 x18. 01528

m = 0.73 gm

mass of vapour = 0.73 g

b) Heat transfer Q = mCp (T₂- T₁)

Q = (900+0. 73) x1x ((90-35)+273)

Q = 295,439 J

Heat transfer from wall, Q = 295.43KJ

c) for barely Warm. coffee

P V = n R T

P = nRT/V

P = 0.04053 x 8.31 X (35 +273) /0.001

P = 1037.856 KPa

pressure at 35^oc or coffee at bare warm stage is = 1037.85 Kpa