A glass of ice water was made by adding six 25 g ice cubes to one liter of water. If the water was initially 25C and the ice cubes were initially at -15C, what is the final temperature of the ice water? (I don't know how to get it.)
See your original post. I answered there.
To determine the final temperature of the ice water, we can use the principle of conservation of energy.
First, let's calculate the amount of heat gained or lost by each component. We can use the formula:
Q = mcΔT
Where:
Q represents the heat gained or lost
m represents the mass
c represents the specific heat capacity of the substance
ΔT represents the change in temperature
Since the specific heat capacity of water is 4.18 J/g°C, and the specific heat capacity of ice is 2.09 J/g°C, we can substitute these values into the formula.
For the water:
Qwater = mwater * cwater * ΔTwater
For the ice:
Qice = mice * cice * ΔTice
Now let's calculate the heat gained or lost by each component.
The initial temperature of the water is 25°C. Since we're assuming there is no energy loss to the surroundings, the heat gained by the ice will equal the heat lost by the water. Therefore:
Qwater = -Qice
To find the masses of water and ice, we have the information that there are six ice cubes, each weighing 25 grams, resulting in a total of 6 * 25 = 150 grams of ice.
The specific heat capacity of water is 4.18 J/g°C, and the initial temperature of the water is 25°C.
Now let's calculate the heat lost by the water:
Qwater = mwater * cwater * ΔTwater
Qwater = 1000 g * 4.18 J/g°C * (Tfinal - 25°C)
Next, let's calculate the heat gained by the ice:
Qice = mice * cice * ΔTice
Qice = 150 g * 2.09 J/g°C * (Tfinal + 15°C)
Since we know that Qwater = -Qice, we can equate the two equations:
1000 g * 4.18 J/g°C * (Tfinal - 25°C) = -150 g * 2.09 J/g°C * (Tfinal + 15°C)
Now solve for Tfinal.