SP1801 DRINK - Drink, on Ice

A good drink is always served on ice. That said, the amount of ice is what makes the difference. If it is too much, the drink will be well cooled, however, this is a bit of fraud as there could be less ice (and more Vodka for example). On the other hand, if there is too little ice the drink is warm which is unacceptable. You are to help the bartender, of course neither with mixing nor drinking, but with calculating the expected outcome of such mixtures. To make things easier, we assume that pure water is mixed with ice in a closed system, i.e., there is no problem with the outside temperature or the warming of the bottle, etc. Therefore, after a some time has passed, the system may be regarded as balanced (there is no further change in temperature and no more melting or freezing). Your job is to calculate the final temperature of this balanced system and the amount of ice and water in this equilibrium state. As you know from physics, it takes 4.19 Joule to heat one gram of water one Kelvin, whereas it takes 2.09 Joule if it is ice. We define the capacities `c $ _{w} $ = 4.19 J/(g*K)` and `c $ _{i} $ = 2.09 J/(g*K)`. Melting one gram of ice takes 335 Joule, where the temperature remains constant at zero. We define the constant `e $ _{m} $ = 335 J/g`. The total thermal energy of the ice and the water before the experiment is equal to the thermal energy of the final mixture. The figure below shows the energy of one gram of ice, ice-water-mixture, or water, where the temperature is measured relative to -30 degrees Celsius. The jump at 0 degrees represents the melting of ice to water. The amount of energy gained is proportional to the amount of ice already melted.

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