If 50.0 kJ of heat are added to a 20.0 gram ice cube at -5.00 degrees Celsius, what will be the resulting state and temperature of the water?
I need some serious help with this! I think I'm making it much harder than it really has to be.
OK. You have 50,000 Joules of heat.
How much heat do you need to move the temperature of the ice at -5.00 C to zero C?
q = mass ice x specific heat ice x (0-(-5)) = ??. Solve for q and subtract joules from 50,000. What's left?
Now how much does it take to melt 20 g ice?
That's q = mass x heat fusion.
Solve for q and see if you have that much heat left from the first part. If so all of the ice will melt. If not, some of it will melt and you can determine how much from the numbers.
Post your work if you get stuck and it would help if you included the heat fusion and the specific heat ice.
All right. I'll go try and work on this problem and come back if I need some help.
The specific heat of ice: 2.092 J/g-Kelvin
The specific heat of water: 4.184 J/g-Kelvin
The specific heat of steam: 1.841 J/g-Kelvin
Heat of Fusion: 6.008 kJ/mol at 0.00 degrees Celsius
Heat of Vaporization: 40.67 kJ/mol at 100 degrees Celsius
Ok, so I'm stuck....Here is my work thus far (please see my previous response for the values of fusion and vaporization, as well as the specific heats)
q=m x specific heat x change in temp
=20.0 g x (2.o92 J/g x Kelvin) x (0--5)=209.2 J
So... 50,000 J - 209.2 J= 49790.8 J left over
so now for the heat of fusion
q= mass x heat fusion= (20.0 g/18g/mol) x 6.006 kJ/mol= 6.675 kJ which is 6675 J
By now, the ice is melted and we still have 43115.8 J left
q=m x specific heat x change in temp for water
=20.0 g x (4.184 J/g x Kelvin) x 100 C=8368 J
This leaves me with 34747.8 J of heat
But if I try to work in the heat of evaporation, I get 45188.8 J and that would leave me with a negative number. I'm not sure where to go from here to find what Temperature my resulting state would be in...
So we start with 50,000 joules heat.
mass ice x specific heat ice x delta T = 20 g x 2.092 J/g x 5 = 209.2 which is much less than 50,000 and all of the ice will move from T = -5 to T = 0 C.
To change ice at zero C to liquid water at zero C we newed
mass ice x heat fusion
20g x 333.5 J/g (I changed 6.008 kJ/mol to J/g) = 6,670 J which means we have enough heat to melt all of the ice. How much heat is left?
50,000 - 209.2 - 6,670 = ??
So the extra J remaining will go to heat the water at zero from the melted ice to some higher T. How much heat is needed to heat the water at zero C to 100 C?
20g x 4.184 J/g*C x 100 = 8368 J.
How much is left now?
50,000 - 209.2-6670-8368 = 34,752.8 J.
ETC.
You were showing your work while I was showing mine so I'll try to catch up. You're right, your last calculation leaves you with a negative number which means that not all of the water can be vaporized. The easy way to know how much CAN be vaporized is to use that same formula
mass water x 2257.56 J/g = 34752.8 ans solve for mass water. I get something like 15 or so.
So 15.394 g H2O x 2257.56 = which checks out.
That will leave 20.00-15.394 = ??g H2O at 100 C and 15.394 g steam at 100 C.
I don't think we are allowed that many significant figures but we can't make sure the heat works out without using all of them. You should round at the end.
I'm working on this same problem
I'm right there with you until the last step where you have Mass Water x 2257.56 = 34752.8
Can you explain to me where the 2257.56 comes from?
THANKS!
No worries, I'm here to help! Let's break down the problem step by step so it becomes easier to understand.
First, we need to determine the amount of heat required to raise the temperature of the ice cube from -5.00 degrees Celsius to the melting point of ice, which is 0 degrees Celsius. To do this, we can use the formula:
Q = m * ΔT * C
Where:
Q is the heat energy
m is the mass of the substance
ΔT is the change in temperature
C is the specific heat capacity of the substance
In this case, the specific heat capacity of ice is 2.09 J/g°C. Therefore, we can calculate the heat energy required using this formula:
Q = (20.0 g) * (0°C - (-5.00°C)) * (2.09 J/g°C)
Simplifying the equation gives us:
Q = (20.0 g) * (5.00°C) * (2.09 J/g°C)
Q = 209 J
So, it will take 209 J of heat energy to raise the temperature of the ice cube from -5.00°C to 0°C.
Next, we need to determine the amount of heat required to melt the ice cube. The heat of fusion, which is the amount of heat required to convert a solid substance into a liquid at its melting point, is 334 J/g for water. Therefore, we can calculate the heat energy required to melt the ice cube using the following formula:
Q = m * ΔHf
Where:
Q is the heat energy
m is the mass of the substance
ΔHf is the heat of fusion
In this case, the mass of the ice cube is 20.0 g. Substituting these values into the equation gives us:
Q = (20.0 g) * (334 J/g)
Q = 6,680 J
So, it will take 6,680 J of heat energy to melt the ice cube completely.
Now, let's determine the total heat energy required. We need to add the heat energy required to raise the temperature of the ice cube to 0°C and the heat energy required to melt the ice cube.
Total heat energy = heat energy to raise to 0°C + heat energy to melt
Total heat energy = 209 J + 6,680 J
Total heat energy = 6,889 J
Since we have a heat input of 50.0 kJ (50,000 J), we can determine how much of the ice will melt and how much heat energy will remain. Subtracting the total heat energy required from the heat input, we get:
Heat energy remaining = Heat input - Total heat energy required
Heat energy remaining = 50,000 J - 6,889 J
Heat energy remaining = 43,111 J
The remaining heat energy will increase the temperature of the water. We can use the formula Q = m * ΔT * C, where the specific heat capacity of water is 4.18 J/g°C, to determine the temperature change:
43,111 J = (20.0 g) * ΔT * (4.18 J/g°C)
ΔT = 43,111 J / (20.0 g * 4.18 J/g°C)
ΔT ≈ 517.10°C
Therefore, the resulting state will be liquid water, and the temperature will increase by approximately 517.10°C.
Note: It's important to note that water cannot reach this temperature without significant increase in pressure due to the change in phase from a solid to a liquid. This problem assumes standard pressure conditions.