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Evaporative cooling of beverages. A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature T=15°C, the environment has temperature Tenv=32°C , and the container is a cylinder with radius r=2.2cm and height 10cm . Approximate the emissivity asε=1, and neglect other energy exchanges. At what rate is the container losing water mass?

Short Answer

Expert verified

The rateat which the container loses water mass is6.8×107kg/s

Step by step solution

01

Identification of given data

i) The temperature of the container with the beverage is T=15°Cor288K

ii) The temperature of the environment is Tenv=32°Cor305K

iii) The radius of the cylinder is r=2.2cmor0.022m

iv) The height of the cylinder is h=10cmor0.10m

v) The emissivity of the material, ε=1

02

Significance of evaporation and thermal radiation

The process of evaporation occurs on the surface of the liquid as it changes from the present liquid state to the gaseous state. Here, a certain amount of heat is released by the body during the phase change process that is radiated as the amount of thermal energy by the system. Thermal radiation is the amount of electromagnetic radiation emitted from a material that is due to the heat radiated by the material depending on its temperature. Thus, the energy lost due to evaporation is equal to the energy gained due to radiation.

Formulae:

The conduction rate at which heat energy is transferred by a body,

Pcond=kA(THTC)L …(i)

where,kis the thermal conductivity of the material,Ais the surface area of radiation,THis the temperature at the hotter end,TLis the temperature at the colder end,Lis the length of the conduction.

The heat energy released by the body, Q=Lvm …(ii)

Where, Lfis the latent heat of fusion, m is the mass of the substance.

The rate of radiation by a body, Prad=σεA(Tenv4T4) …(iii)

Where,σis the Stefan–Boltzmann constant and is equal to(5.67×108W/m2.K4),ϵis the emissivity of the substance,Ais the surface of the area of radiation,Tenvis the temperature of the environment,Tis the thermodynamic temperature.

03

Determining the rate at which the container losses water

Energy is lost by the beverage due to evaporation. Then, the heat of vaporization is given by using equation (ii), where

The heat of vaporization,LV=2.256×106J/kg

Differentiating equation (ii) with respect to time, we get the rate as:

dQdt=LVdmdt …(iv)

The energy lost due to evaporation equals the energy gained by radiation. Hence, equating equation (iv) with equation (iii), we get that

LVdmdt=σεA(Tenv4T4) …(v)

The total surface area of the container (a cylinder with one end open) is given as:

A=πr2+2πrh=3.14(0.022m)2+2×3.14×0.022m×0.10m=1.53×102m2

Substituting the given values and the above area, equation (v) gives the rate of mass loss as:

dmdt=σεA(Tenv4T4)LV=5.67×108W/m2.K4×1.0×1.53×102m2×((305K)4(288K)4)2.256×106J/kg=6.8×107kg/s

Hence, the required value of the mass loss by the container is 6.8×107kg/s

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