You are a member of an oceanographic group that is studying living on the ocean
ID: 726157 • Letter: Y
Question
You are a member of an oceanographic group that is studying living on the ocean floor. The level at which you are planning to live has a pressure of 1. 465 atm; therefore, you don't have to use an abode that is maintained at 1 atm. However, you want to boil semi water to make coffee. If your coffee loses heat at a rate of 10. 246 kJ/min, how long would you have to let it cool, after boiling the water and making coffee, so that the coffee were at the drinkable temperature of 60 degree C? Heat of vaporization of water Delta Hvap = 40. 7 kJ/mol, assume that you have 400 mL of coffee, the density of the coffee is the same as water (0. 9679700 g/cm3), and the heat capacity of the coffee is the same as water (4. 184 J/g. K).Explanation / Answer
The trick here is that the pressure is greater, so that the water is above 100C when it boils. This is foods with water have high altitude recipes where you need to cook longer; there the boiling point is lower. This is also why pressure cookers take less time to cook food. The Clausius-Clapeyron equation can be used Normal boiling point = ((R ln(Po)/delta H vap) + 1/To))^-1 Normal boiling point is 273.15 K When the water boils at the higher pressure Po = 1.465 atm R = 0.008314 kJ/(Kelvin*mol) We can solve this for To, which is our new boiling point I solved this iteratively on Excel 373.15 = (7.8E-5 + 1/384.34)^-1 So the new boiling point is 384.34 Now, to bring this down to 60 C or 273.15 + 60 = 333.15K is a drop of 51.19 degrees. We have slightly less than 400 g water 400ml*0.96796g/ml = 387.2g m cp delta t = 387.2g*4.184J/gK*51.19K = 82,927 J This is the heat we need to lose 82,927J/10,246J/min = 8.09 minutes 8 minutes 6 seconds
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