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 $\mathit{T}\mathbf{=}\mathbf{}\mathbf{15}\mathbf{\text{°C}}$, the environment has temperature ${\mathit{T}}_{\mathbf{e}\mathbf{n}\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{32}\mathbf{\text{°C}}$ , and the container is a cylinder with radius $\mathit{r}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.2}\mathbf{}\mathbf{\text{cm}}$ and height $\mathbf{10}\mathbf{}\mathbf{\text{cm}}$ . Approximate the emissivity as$\mathit{\epsilon }\mathbf{=}\mathbf{}\mathbf{1}$, and neglect other energy exchanges. At what rate is the container losing water mass?

i) The temperature of the container with the beverage is $T=15°\text{Cor288K}$

ii) The temperature of the environment is $T_{env}=32°\text{Cor305K}$

iii) The radius of the cylinder is $r=2.2\text{cmor0.022m}$

iv) The height of the cylinder is $h=10\text{cmor0.10m}$

v) The emissivity of the material, $\epsilon =1$

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,

$P_{cond}=\frac{kA(T_H-T_C)}{L}$ …(i)

where,$k$is the thermal conductivity of the material,$A$is the surface area of radiation,$T_H$is the temperature at the hotter end,$T_L$is the temperature at the colder end,$L$is the length of the conduction.

The heat energy released by the body, $Q=L_{v}m$ …(ii)

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

The rate of radiation by a body, $P_{rad}=\sigma \epsilon A(T_{env}^4-T^4)$ …(iii)

Where,$\sigma$is the Stefan–Boltzmann constant and is equal to$(5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}{\text{.K}}^4)$,$ϵ$is the emissivity of the substance,$A$is the surface of the area of radiation,$T_{env}$is the temperature of the environment,$T$is the thermodynamic temperature.

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,$L_V=2.256\times {10}^6\text{J/kg}$

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

$\frac{dQ}{dt}=L_V\frac{dm}{dt}$ …(iv)

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

$L_V\frac{dm}{dt}=\sigma \epsilon A(T_{env}^4-T^4)$ …(v)

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

$\begin{array}{c}A=\pi r^2+2\pi rh\\ =3.14{(0.022\text{m})}^2+2\times 3.14\times 0.022\text{m}\times 0.10\text{m}\\ =1.53\times {10}^{-2}{\text{m}}^2\end{array}$

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

$\begin{array}{c}\frac{dm}{dt}=\frac{\sigma \epsilon A(T_{env}^4-T^4)}{L_V}\\ =\frac{5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}{\text{.K}}^{\text{4}}\times 1.0\times 1.53\times {10}^{-2}{\text{m}}^{\text{2}}\times ({(305\text{K})}^4-{(288\text{K})}^4)}{2.256\times {10}^6\text{J/kg}}\\ =6.8\times {10}^{-7}\text{kg/s}\end{array}$

Hence, the required value of the mass loss by the container is $6.8\times {10}^{-7}\text{kg/s}$