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 matched the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature `T = 10^(@)C`, the environment has temperature `T_("env") = 32^(@)C`, and the container is a cylinder with radius `r = 2.2 cm` and height 10cm. Approximate the emissivity as `epsilon =1`, and neglect other energy exchanges .At what rate dm`//` dt is the container losing water mass ?

Answer

To solve the problem of evaporative cooling of beverages in a porous ceramic container, we will follow these steps:

### Step 1: Understand the Energy Exchange
The energy lost due to evaporation of water from the container equals the net energy gained via radiation from the environment. We can express this mathematically as:
\[ Q_{\text{evap}} = Q_{\text{rad}} \]

### Step 2: Calculate the Surface Area
The container is a cylinder, and we need to calculate the area from which energy exchange occurs. The total area \( A \) consists of the top area and the side area:
- Top area: \( A_{\text{top}} = \pi r^2 \)
- Side area: \( A_{\text{side}} = 2 \pi r h \)

Given:
- Radius \( r = 2.2 \, \text{cm} = 0.022 \, \text{m} \)
- Height \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)

Calculating the areas:
\[ A_{\text{top}} = \pi (0.022)^2 = \pi \times 0.000484 = 0.00152 \, \text{m}^2 \]
\[ A_{\text{side}} = 2 \pi (0.022)(0.1) = 0.0044 \, \text{m}^2 \]
Thus, the total area \( A \):
\[ A = A_{\text{top}} + A_{\text{side}} = 0.00152 + 0.0044 = 0.00592 \, \text{m}^2 \]

### Step 3: Apply the Stefan-Boltzmann Law
The energy gained via radiation can be expressed using the Stefan-Boltzmann law:
\[ Q_{\text{rad}} = \epsilon A \sigma (T_{\text{env}}^4 - T^4) \]
Where:
- \( \epsilon = 1 \) (emissivity)
- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) (Stefan-Boltzmann constant)
- \( T_{\text{env}} = 32^\circ C = 305 \, \text{K} \)
- \( T = 10^\circ C = 283 \, \text{K} \)

Calculating \( Q_{\text{rad}} \):
\[ Q_{\text{rad}} = 1 \times 0.00592 \times 5.67 \times 10^{-8} \times (305^4 - 283^4) \]
Calculating \( 305^4 \) and \( 283^4 \):
\[ 305^4 \approx 8.617 \times 10^9 \quad \text{and} \quad 283^4 \approx 6.469 \times 10^9 \]
Thus,
\[ Q_{\text{rad}} = 0.00592 \times 5.67 \times 10^{-8} \times (8.617 \times 10^9 - 6.469 \times 10^9) \]
\[ = 0.00592 \times 5.67 \times 10^{-8} \times 2.148 \times 10^9 \]
Calculating this gives:
\[ Q_{\text{rad}} \approx 0.00592 \times 5.67 \times 2.148 \times 10^1 \approx 0.000071 \, \text{W} \]

### Step 4: Relate Energy Loss to Mass Loss
The energy lost due to evaporation can be expressed as:
\[ Q_{\text{evap}} = \frac{dm}{dt} L \]
Where \( L \) is the latent heat of vaporization of water:
- \( L = 2.265 \times 10^6 \, \text{J/kg} \)

Setting \( Q_{\text{evap}} = Q_{\text{rad}} \):
\[ \frac{dm}{dt} L = Q_{\text{rad}} \]
\[ \frac{dm}{dt} = \frac{Q_{\text{rad}}}{L} \]
Substituting the values:
\[ \frac{dm}{dt} = \frac{0.000071}{2.265 \times 10^6} \approx 3.13 \times 10^{-8} \, \text{kg/s} \]

### Step 5: Convert to Milligrams per Second
To convert \( \frac{dm}{dt} \) to milligrams per second:
\[ \frac{dm}{dt} \approx 3.13 \times 10^{-8} \, \text{kg/s} \times 10^6 \, \text{mg/kg} \approx 0.0313 \, \text{mg/s} \]

### Final Answer
The rate at which the container is losing water mass is approximately:
\[ \frac{dm}{dt} \approx 0.0313 \, \text{mg/s} \]