Question

A glass coffee pot has a 9.00 cm diameter circular bottom in contact with a heating element that keeps the coffee warm with a continuous 50.0-W heat input.

(a) What is the temperature of the bottom of the pot, if it is 2.00 mm thick and the inside temperature is 57.8°C? The thermal conductivity of the pot is 0.84 J/(s · m · °C).
°C

(b) If the temperature of the coffee remains constant and all of the heat input is removed by evaporation, how many grams per minute evaporate? Take the latent heat of vaporization to be 560 cal/g.
g

Answer 1

a)

Power transferred , P = k*A*(Tf - Ti)/t

where k = thermal conductivity = 0.84 J/s.m.C

A = area = pi*(0.09^2)/4

(Tf) = Temperature of bottom

Ti = Temperature of top

t = thickness = 2 mm = 0.002 m

So, 50 = 0.84*(pi*0.09^2/4)*(Tf - 57.8)/0.002

So, Tf = 76.5 deg C <--------answer

b)

Latent heat of vaporization = 560 cal/g = 560*4.18 J/g

Heat transferred = 50W = 50 J/s = 50*60 = 3000 J/min

So, weight of coffee evaporating = 3000/(560*4.18) = 1.28 g/min <-------answer