An ice cube at 0.00 degree celsius with a mass of 23.5 g is placed into 550.0 g of water, initially at 28.0 degree celsius, in an insulated container. Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all the ice has melted?
Close but not quite right. I see you have used kJ/mol for heat fusion but J/g for specific heat water. Units must be consistent. I think the heat fusion is 333 J/g but check me out on that. For the mass of the melted ice don't you think that 23.5 g solid ice will produce 23.5 g liquid water (and 23.5 g steam as far as that goes)?
heat to melt ice + heat to raise T of melted ice from zero to final T - heat lost by water at 28 in falling to final T.
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass warm water x specific heat water x (Tfinal-Tinitial)] = 0
There is only one unknown in this equation. That is Tfinal. Solve for that. I think it is easier (far easier) to substitute the numbers and work it through that way rather than try to solve algebraically for Tfinal.
thanks for the help. im still a bit confused.,,
[23.5g x 6.02] + [? x 2.09 x (Tfinal- 0)] + [550.g x 4.184 x (Tfinal- 28.0)] = 0
is this right so far? How would you be able to get the mass of melted ice?
To find the final temperature of the entire water sample after all the ice has melted, we need to use the principle of conservation of energy and apply the equation for heat transfer.
First, let's calculate the energy required to melt the ice. This can be done using the equation:
Q = m * L
Where:
Q = Energy (in Joules)
m = Mass of ice (in grams)
L = Latent heat of fusion for water (in J/g)
The latent heat of fusion for water is the amount of energy required to convert one gram of ice to water at its melting point. For water, the value is 334 J/g.
Thus, the energy required to melt the ice can be calculated as:
Q1 = 23.5 g * 334 J/g = 7839 J
Next, let's determine the energy transferred from the water to the ice to melt it. This can be calculated using the equation:
Q2 = m * c * ΔT
Where:
Q2 = Energy transferred (in Joules)
m = Mass of water (in grams)
c = Specific heat capacity of water (in J/g°C)
ΔT = Change in temperature of the water (in °C)
The specific heat capacity of water is approximately 4.18 J/g°C.
First, we need to calculate the change in temperature of the water. Since the ice starts at 0.00°C and melts completely, its temperature remains constant at 0.00°C until it melts. The resulting water will then be at 0.00°C.
Therefore, the change in temperature is:
ΔT = 0.00°C - 28.0°C = -28.0°C
Next, let's calculate the energy transferred:
Q2 = 550.0 g * 4.18 J/g°C * -28.0°C = -64676 J
Now, let's find the total energy exchanged:
Q_total = Q1 + Q2
= 7839 J + (-64676 J)
= -56837 J
The negative sign indicates that energy has been lost by the system due to the heat transferred from the water to the ice.
Now, let's find the final temperature of the entire water sample.
Since no heat is lost to the surroundings, the total energy exchanged must be zero. Therefore, we can write:
Q_total = m_total * c * ΔT
Where:
m_total = Total mass of the water sample (including the melted ice)
c = Specific heat capacity of water
ΔT = Change in temperature of the water (final temperature - initial temperature)
Substituting the values:
-56837 J = (550.0 g + 23.5 g) * 4.18 J/g°C * ΔT
Now we can solve for ΔT:
ΔT = -56837 J / ((550.0 g + 23.5 g) * 4.18 J/g°C)
≈ -0.199°C
Since the ice has melted completely and the resulting water is at 0.00°C, the final temperature of the entire water sample is approximately:
Final temperature = 28.0°C + ΔT
≈ 28.0°C - 0.199°C
≈ 27.80°C
Therefore, the temperature of the entire water sample after all the ice has melted is approximately 27.80°C.
I asked my professor and she said when the ice cube melts, it absorbs energy from the water. We can figure out how much energy using the heat of fusion. The water cools down; since we know how much energy it lost, how much mass it has (including the mass of the ice cube, now part of the water), and the specific heat of water we can calculate the temperature change using the heat equation from Ch 3 (q = mass x delta T x specific heat).
So what I did was 23.5 g x 1mol/18.02g x 6.02kj/1 mol =7.850 kj
Then Plugged into the equation of q=mCdeltat 7850j= 573.5 g x 4.184 j/g x (x-28) and from
That I solved for x = 31.27 but that doesn't make any sense since it should cool down... The answer in the book is 23.6...