Use standard enthalpies of formation to calculate the standard
ch...

Question

Use standard enthalpies of formation to calculate the standard
change in enthalpy for the melting of ice. (The ΔH °f for H2O(s)
is -291.8 kJ/mol.) Use this value to calculate the mass of ice
required to cool 355 mL of a beverage from room temperature
(25.0 °C) to 0.0 °C. Assume that the specific heat capacity and
density of the beverage are the same as those of water.

Accepted Answer

Hey everyone were asked how much ice in grams is needed to decrease the temperature of 650 mL of milk. And they provided us the standard entropy of formation of our ice and our density of milk and its specific heat. And this is from 75 degrees Celsius to 15 degrees Celsius first. Let's go ahead and conceptualize this. So we have our ice and this is going to be water in its solid state and this is going to turn into liquid water. So our standard entropy for our eyes, which is water in its solid state, was said to be negative 2 91.8 kg joules per mole. And if we look at our text books, the standard envelop of our water in its liquid state comes up to negative 285.8 kg joules per mole. So the change in entropy of our reaction is going to be water in its liquid state minus water in its solid state. And when we plug in our values we have negative 285.8 kg joules per mole. And we're going to subtract negative 291. kg joules per mole. And this gets us to six kg joules per mole. Now let's go ahead and determine the mass of our milk. Using our 650 ml of milk, we're going to use dimensional analysis and we know that we have 10 to third milliliters per one leader. Now using the density that they provided us with, we know that we have 1.35 kg per one liter. And lastly we want to convert those kilograms into grams. And we know that per one kg we have 10 to the third grams. So when we calculate this out and cancel out all of our units, we end up with a mass of 672.75 g of milk. Next we want to determine the heat needed to cool our milk and we can do so by using the Formula Q equals mass times our specific heat times our change in temperature. So plugging in these values, we have 672.75 g. And we were told we had a specific heat of 3. jules over grams times degree Celsius. And our change in temperature Was said to be 15.2°C -75°C. And we want to convert this into killing jewels. So we know that we have one. Kill a jewel and this will contain 10 to the third jewels. So when we calculate this out and cancel out all of our respective units, we end up with the heat of negative 157.1985 kg jewels. Lastly, we want to calculate the mass of our ice. So using that heat of 157.01985 kg jewels. We're going to use dimensional analysis and we're going to take our change in entropy and we know that we had six kg joules per one mole. And lastly, we want to use our molar mass of water, which we know to be 18.01 grams per one mole, calculating this out. We end up with 471g of our ice, and this is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.

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