During a hot summer day, a \(2-L\) bottle drink is to be cooled by wrapping it in a cloth kept wet continually and blowing air to it with a fan. If the environment conditions are \(1 \mathrm{~atm}, 80^{\circ} \mathrm{F}\), and 30 percent relative humidity, determine the temperature of the drink when steady conditions are reached.

First, we need to find the wet-bulb temperature to determine the temperature to which the drink will be cooled. The wet-bulb temperature can be found using a psychrometric chart or equations. For simplicity, we will use the following empirical formula to estimate the wet-bulb temperature in Celsius: \(T_{wb} = T_a - 0.56(1 - RH)(T_a - 14.3) \) Where \(T_{wb}\) is the wet-bulb temperature in Celsius, \(T_a\) is the ambient temperature in Celsius, and \(RH\) is the relative humidity in decimals. Convert the given ambient temperature from Fahrenheit to Celsius: \(T_a (^{\circ}C) = \frac{80 - 32}{1.8} = 26.67 ^{\circ}C\) Now, we can calculate the wet-bulb temperature: \(T_{wb} = 26.67 - 0.56(1 - 0.3)(26.67 - 14.3) \) \(T_{wb} = 26.67 - 0.56(0.7)(12.37) \) \(T_{wb} ≈ 22.7 ^{\circ}C \)

Now that we have the wet-bulb temperature in Celsius, we need to convert it back to Fahrenheit so that we can express the steady-state temperature of the drink in the original unit: \(T_{drink} (^{\circ}F) = 1.8(T_{wb}) + 32\) \(T_{drink} (^{\circ}F) = 1.8(22.7) + 32\) \(T_{drink} ≈ 72.9 ^{\circ}F\)

The steady-state temperature of the 2-L bottle drink will be the wet-bulb temperature since the evaporative cooling process will cool the drink down to that temperature. Therefore, the drink will reach a temperature of: \(T_{drink} ≈ 72.9 ^{\circ}F\) When steady conditions are reached, the temperature of the drink will be approximately 72.9°F.