Assuming that Coca-Cola has the same specific heat as water, how much energy in calories is removed when 350 g of Coca-Cola (about the contents of one 12 oz can) is cooled from room temperature ( 25 ° C ) to refrigerator temperature ( 3 ° C )?

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Hello, everyone. Today we are being given the following problem. Ethyl acetate is a solvent use for rec rec crystallization processes. Ethyl acetate has a specific heat of 0. calories per grams times a degree Celsius. Calculate the heat energy lost in the cooling process. When 111 g of ethyl acetate was heated to its boiling 0.77 0.1 degrees Celsius to dissolve a sample and then cool down to room temperature 25 degrees Celsius assume that the specific heat does not change. So first, we need to pull out our equation for heat, which says that Q or heat is equal to the mass times our specific heat capacity times our change in temperature, which is going to be final temperature minus initial temperature. So far at mass, we had 111 g. Our specific heat capacity is 0. calories per grams times degrees Celsius. And then our change in temperature, it was cool to 25 degrees Celsius, but it was initially heated to 77.1 degrees Celsius. And so when we plug this into our calculator, and our units canceled out we are left with negative 2. times 10 to the third calories as our final answer. And with that, we have solved the problem overall. I hope this helped. And until next time.

Assuming that Coca-Cola has the same specific heat as water, how much energy in calories is removed when 350 g of Coca-Cola (about the contents of one 12 oz can) is cooled from room temperature ( 25 ° C ) to refrigerator temperature ( 3 ° C )?

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Key Concepts

Specific Heat Capacity

Heat Transfer

Temperature Change