Answer:
See explanation below
Explanation:
First, you are not providing any data about the mass of ice and the innitial temperature of water, so, I'm gonna use data from a similar exercise, so in order for you to get the accurate and correct answer, just replace the data in this procedure, and you should be fine.
Now, For this exercise, I will assume we have an 8 g ice cube pounded into 230 g of water. The expression to use here is the following:
q = m*Cp*ΔT (1)
Solving for ΔT:
ΔT = q / m*Cp (2)
Where:
q: heat of the water
m: mass of water
Cp: specific heat of water which is 4.18 J /g °C
Now, we don't know the heat emmited by water, we need to calculate that. To do this, we have data for ice and water, so, let's find first the heat absorbed by the melting ice, and then, the water.
Converting the grams into moles, using the molar mass of water which is 18 g/mol rounded:
moles of ice = 8 g / 18 g/mol = 0.44 moles of water
Now, we'll use the molar heat of fusion of water to convert the moles to kJ:
qi = 0.44 mol * 6.02 kJ/mol =¨2.6488 kJ
Now, as the ice is the system and water the surroundings, the melting of ice in endothermic therefore the heat of water should be the same of ice but negative, therefore:
qi = -qw = -2.6488 kJ or -2648.8 J
Finally, replace this value in equation (2) to get the temperature change:
ΔT = -2648.8 / (230 * 4.18)
ΔT = -2.75 °C
Now, use your data of ice and water and replace them here in this procedure to get the correct and accurate answer.
