If 5.40 kcal of heat is added to 1.00 kg of ice at 0⁰C, how much liquid water at 0⁰C is produced? Show all calculations leading to an answer.
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To solve this problem, we need to consider the heat required to convert the ice at 0⁰C into liquid water at 0⁰C.
The specific heat capacity of ice is given as 2.09 J/g·⁰C, which means it takes 2.09 Joules of energy to raise the temperature of 1 gram of ice by 1⁰C.
First, let's convert the 1.00 kg of ice into grams:
1.00 kg = 1000 grams
Now, let's calculate the heat required to raise the temperature of the ice from 0⁰C to its melting point (0⁰C):
Q1 = mass × specific heat capacity × temperature change
= 1000 g × 2.09 J/g·⁰C × (0⁰C - 0⁰C) // The temperature change is 0
= 0 J
As the temperature is not changing, no heat is required to raise the ice to its melting point.
Next, we need to calculate the heat required to melt the ice completely into liquid water at 0⁰C. The heat of fusion for water is 334 J/g, which means it takes 334 Joules of energy to convert 1 gram of ice into liquid water at 0⁰C.
Q2 = mass × heat of fusion
= 1000 g × 334 J/g
= 334,000 J
Now that we have calculated the total heat required to melt the ice, let's convert it into kcal:
1 kcal = 4184 J // This is the conversion factor from Joules to kcal
Q2 (in kcal) = 334,000 J × (1 kcal/4184 J)
= 79.86 kcal
Therefore, a total of 79.86 kcal of heat is required to melt 1.00 kg of ice completely into liquid water at 0⁰C.
Since we are adding 5.40 kcal of heat, we can calculate the amount of liquid water produced by subtracting the heat used for melting from the total heat added:
Heat for melting = 79.86 kcal
Heat added = 5.40 kcal
Liquid water produced = Heat added - Heat for melting
= 5.40 kcal - 79.86 kcal
= -74.46 kcal
The negative value indicates that the heat added is not sufficient to melt the ice completely into liquid water. Therefore, no liquid water would be produced.