(Solved): Problem 1: Specific heat and evaporation The specific heat value of a substance is the amount of en ...

Problem 1: Specific heat and evaporation The specific heat value of a substance is the amount of energy required to change the temperature of \( 1 \mathrm{~kg} \) of the material by \( 1 \mathrm{~K} \). Water has a fairly high specific heat; compared to other materials: \( 1 \mathrm{~kg} \) of water must absorb (or release) \( 4187 \mathrm{~J} \) of energy to change its temperature by \( 1 \mathrm{~K} \). A. Most' people like drinking hot drinks such as coffee at about \( 60^{\circ} \mathrm{C} \). However they are usually served at about \( 80^{\circ} \mathrm{C} \). Say you have \( 180 \mathrm{~mL} \) of a water-based hot drink (same specific heat as water). 1. If the hot drink is in a sealed container, how much heat must it loose before it is at a comfortable drinking temperature? This is a fairly slow process, due to the low thermal conductivity of the material of the cups. To speed up the cooling you could instead leave the lid off your drink so that some of it can evaporate. Each molecule that evaporates must gain energy to break the hydrogen bonds with all of its neighboring molecules. They receive this energy from the rest of the liquid, decreasing the internal energy of the remaining liquid and thus cooling it. The energy acquired by the evaporating molecules (or loss by the liquid) is called the heat of vaporization. For water it is \( 40660 \frac{1}{\mathrm{~mol}^{\circ}} \) or \( 6.75 \times 10^{-20} \frac{1}{\text { molecule }^{-}} \) B. Assume that \( 9 \mathrm{~mL} \) of your hot drink evaporate (assume its molar mass is that of water). 1. How much energy was removed from the liquid to evaporate that amount? 2. Calculate the final temperature of the drink. Important: since the amount of liquid changes, you can't use the specific heat to find its energy change. Instead recall that the internal energy of water is \( U=6 N k_{B} T=6 n R T \), in terms of the changing number \( N \) of molecules or \( n \) of moles. Determine \( n_{i} \) and \( n_{f} \) and find an expression for \( \Delta U=U_{f}-U_{i} \). Use it to calculate \( T_{f} \). C. We would like to know how much of the drink must evaporate for it to reach the drinkable temperature (we want to solve for the final number particles \( N_{f} \), or moles \( n_{f} \) ). 1. Write an expression for \( \Delta U=U_{f}-U_{i} \) in terms of \( T_{i}, T_{f}, N_{i} \), and \( N_{f} \) (or better moles, \( n_{i} \), and \( n_{f} \) ) [Hint: it is best to plug values a bit later in the process. Only write down the expression here.] 2. Just like before, write now an expression for the amount of energy carried away by the evaporating molecules (in terms of the heat of vaporization, \( N_{i} \), and \( N_{f} \) ). How does this energy relate to the change of energy \( \Delta U \) of the remaining liquid? (Hint: pay special attention to the signs of each energy change). Use the two expressions for \( \Delta U \) (with the correct signs) to find \( N_{f} \) - 3. Determine the percent of the drink that was lost, \( \frac{N_{t}-N_{f}}{N_{i}} \) or \( \frac{n_{i}-n_{f}}{n_{i}} \) (they are the same). D. This method of cooling down a drink still involves waiting a while. If we're impatient, we could instead add a little tap water (say at \( 15^{\circ} \mathrm{C} \) ). Find the volume of tap water necessary to cool the hot drink to \( 60^{\circ} \mathrm{C} \), assuming no energy is lost to either evaporation or conduction.