Question

During a hot summer day when the air temperature is \(35^{\circ} \mathrm{C}\) and the relative humidity is 70 percent, you buy a supposedly "cold" canned drink from a store. The store owner claims that the temperature of the drink is below \(10^{\circ} \mathrm{C}\). Yet the drink does not feel so cold and you are skeptical since you notice no condensation forming outside the can. Can the store owner be telling the truth?

Short answer

$$e_s(35) \approx 56.548 \, \mathrm{mb}$$ #tag_title#Step 2: Calculate the actual vapor pressure#tag_content#Now, we need to determine the actual vapor pressure, which is directly related to the relative humidity. We can use the following formula: $$e = \frac{RH \times e_s}{100}$$ where \(e\) is the actual vapor pressure, \(RH\) is the relative humidity (35% in this case), and \(e_s\) is the saturation vapor pressure we calculated in step 1. Let's find the actual vapor pressure. $$e = \frac{35 \times 56.548}{100}$$ $$e \approx 19.792 \, \mathrm{mb}$$ #tag_title#Step 3: Calculate the dew point#tag_content#Now, we will calculate the dew point of the air outside by using the following formula: $$T_d = \frac{243.04 \times \ln(e/6.1094)}{17.625 - \ln(e/6.1094)}$$ where \(T_d\) is the dew point in degrees Celsius, and \(e\) is the actual vapor pressure. Let's find the dew point. $$T_d = \frac{243.04 \times \ln(19.792/6.1094)}{17.625 - \ln(19.792/6.1094)}$$ $$T_d \approx 13.673^{\circ}\, \mathrm{C}$$ #tag_title#Step 4: Analyze the dew point in relation to the claimed temperature of the drink#tag_content#Now that we have the dew point (13.673°C), let's compare it to the claimed temperature of the drink (10°C). Since the dew point (13.673°C) is higher than the claimed temperature of the drink (10°C), condensation should form on the can according to the given conditions. If there is no condensation, then the drink might be warmer than the claimed temperature.

Step-by-step solution

Calculate the saturation vapor pressure outside the can

First, we need to calculate the saturation vapor pressure outside the can at the given air temperature. We can use the August-Roche-Magnus approximation to calculate the saturation vapor pressure in millibars (mb): $$e_s(T) = 6.1094 \times e^{\left(\frac{17.625 \times T}{243.04 + T}\right)}$$ where \(T\) is the temperature in degrees Celsius. In this case, the air temperature is \(35^{\circ} \mathrm{C}\). Let's find the saturation vapor pressure. $$e_s(35) = 6.1094 \times e^{\left(\frac{17.625 \times 35}{243.04 + 35}\right)}$$