If 142.38 g of l water at 21.7 degrees C is placed into a syrofoam cup with 174.36 g of ice at -27.5 degrees C what would be the final temperature of the entire contents at equilibrium? If it is partially frozen, how many g of the ice has melted?
given values
cp water = 4.184 j/g degree c
cp ice = 2.114 j/g degree c
heat of fusion of ice is 335 J/g
so far, i have this:
(174.36g)(2.114g)(27.5 C)
= 10247 g to three sig figs = 10200 J
is that i calculate the energy required to raise the ice to 0 degrees C?
now i know that the next steps are to find the enrgy required to melt the ice and the energy reuired to cool the water down to 0 degrees C. how do i do those two calculations?
after that i did this to try and melt the ice
q = (mass l H20/molar mass H20)X(335 j/g) the 335 is the heat of fusion.
is that correct? if not, what am i supposed to do?
Your 10,247 J I get 10,136 J. Check my arithmetic. THEN, I would keep all the numbers and round at the end; otherwise, it may lead to rounding errors. I'll assume the 10,136 J is correct to move the ice from -27.5 to Oo C.
Then we need to calculate the energy required to cool the liquid water from 21.7 C to zero C. That is
142.38 x 4.184 x 21.7 = check me on this but I get about 13,000 J or so. You do the exact number.
So if you look at the numbers, we can move the ice to zero and move the lqiuid water to zero and have about 3,000 J left in the liquid water. That will melt some of the ice. How much?
q = mass x 335 J/g. Calculate the mass of ice that will melt.
I hope this helps. Check my thinking. Check my work.
To find the final temperature of the entire contents at equilibrium, you need to consider the heat transfer that occurs during the process.
First, calculate the energy required to raise the ice from -27.5°C to 0°C. You correctly used the formula:
q1 = (mass of ice) × (specific heat capacity of ice) × (change in temperature)
q1 = (174.36 g) × (2.114 J/g°C) × (0°C - (-27.5°C))
q1 = 10246.83 J (rounded to four significant figures)
So, the energy required to raise the ice to 0°C is approximately 10246.8 J.
Next, you need to calculate the energy required to melt the ice at 0°C. You can use the formula:
q2 = (mass of ice melted) × (heat of fusion of ice)
Since you want to find the amount of ice melted, you can rearrange the formula:
(mass of ice melted) = q2 / (heat of fusion of ice)
Substituting the values, you get:
(mass of ice melted) = 10246.8 J / 335 J/g
(mass of ice melted) ≈ 30.68 g (rounded to two decimal places)
So, approximately 30.68 g of ice has melted.
Finally, you need to calculate the energy required to cool the water from 21.7°C to 0°C. You can use the formula:
q3 = (mass of water) × (specific heat capacity of water) × (change in temperature)
q3 = (142.38 g) × (4.184 J/g°C) × (0°C - 21.7°C)
q3 ≈ -12926.609 J (rounded to four decimal places)
Note: The negative sign indicates that heat is lost from the water.
The total energy change is given by:
Total energy change = q1 + q2 + q3
Total energy change ≈ 10246.8 J + q2 - 12926.609 J
To reach equilibrium, the total energy change should be zero:
10246.8 J + q2 - 12926.609 J = 0
Now, solve for q2 (the energy required to melt the ice):
q2 ≈ 12926.609 J - 10246.8 J
q2 ≈ 2680.809 J (rounded to three decimal places)
Note: This value represents the energy released by the water when it cools down.
To find the final temperature of the entire contents at equilibrium, you need to consider the heat lost by the water (q3) and the heat gained by the melted ice (q2). You can use the equation:
q3 = q2
Substituting the values:
-12926.609 J = 2680.809 J
Since the absolute value of the heat loss and gain are not equal, the equality is not satisfied, and equilibrium is not reached. This means the system will continue to lose heat until it reaches equilibrium.