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.

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Welcome back everyone to another video use standard piece of formation to calculate the standard change in lb for the melting of ice. The standard entropy affirmation of ice is negative 291.8 kilojoules per mole. Use this value to calculate the mass of ice required to cool 355 mL of a beverage from room temperature or 25 °C to zero °C assume that the specific heat capacity and density of the beverage are the same as those of water. And we are given for answer choices. Now A is 121.3 g. B 111.5 g. C 123.5 g and D 119.8 g. So first of all, let's understand the process of melting. We understand that if we take solid water, which is ice, we are going to include H2O solid, it essentially should be converted into liquid water, right. So we're going to draw equilibrium arrows and also state H2O liquid on the right side of the equation. Now, for this process, the end will be changed would be equal to the change of formation of products. In this case, that would be water, liquid minus the entropy change of formation of the reactions. So that would be h2o solid or basically ice. Now, first of all, we need to use the table to identify the first value according to the tables, that'll be negative 285.8 kilojoules per mole. And we need to subtract the value given to us in the problem negative 291.8 kilojoules per mole. Now the result that we get here is 6.0 kilojoules per month. Well, then, so now that we have the entropy change of the process, we need to continue the problem and we want to calculate the amount of heat to, required to cool down the beverage or basically how much heat should we remove from the hot beverage? And we are going to use the formula CM delta C. According to the problem, we're going to use the specific heat capacity of water. We can say that this is 4.184 joules per gram per Celsius. What would be the mass? Well, if the density of water is 1 g per milliliter, then we're going to use 355 g and the temperature change going from 25 to 0, right? That would be negative 25. Now we can calculate Q and we get negative 37,133 joules. So this is the amount of heat that we need to remove from the hot beverage to essentially cool it down, right heat is removed. Hence, we get that negative sign. However, we have to remember that according to the law of energy conservation Q of the reaction should be equal to the negative Q or basically the opposite value to the previous result, which is 37,133 joules. So this is the amount of heat that the ice cube should absorb from here. We also need to remember that cue of the reaction, right? For the melting process is equal to the entropy of the reaction multiplied by mass. OK. So if we solve for mass, we know that mass is equal to Q of the reaction divided by the entropy change. We're going to take 37,133 Jews. And we're going to divide that by six kilojoules per mole, six kilojoules that would be six multiplied by 10 to the third joules because we want to make sure that our units are consistent per mole. And what we notice is that we get moles, right? So if we want to convert that into grams, we also need to multiply by 18.015 g per mole, which is the molar mass of water or basically ice as well. And now for this calculation, we get our final result, that would be 111.5 g. According to the answer choices that we're given, we can state that the correct answer to this problem is B 111.5 g. Thank you for watching.

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Key Concepts

Enthalpy of Formation

Specific Heat Capacity

Mass and Density Relationship