A 500 mL bottle at room temperature (25˚C) of water is poured over 120 g of ice that is (-8˚C). What will be the final temperature of the water be when all the ice has melted? You can ignore loss of heat to the room.
q1 = heat to raise T of solid ice from -8 to zero C.
q1 = mass ice x specific ice x (Tf - Ti) where Tf is 0 C and Ti is -8 C.
q2 = heat to melt ice at zero C to liquid H2O at zero C.
q2 = mass ice x heat fusion
q3 = heat to raise water from zero C to final T.
q3 = mass H2O x specific heat H2O x (Tf - Ti). Ti is zero.
q4 = heat lost by the 500 mL H2O at 25 C.
q4 = mass H2O x specific heat H2O x (Tf - Ti). Ti = 25 C.
q1 + q2 + q3 + q4 = 0
Plug all of that into q1 through q4 and solve for T. Tf is close to 18 C.
Dr. Bob,
When I use the values that I have for latent heat of fusion and vaporization, I do not get the right answer.
I have that latent heat of vaporization is 597.3 - 0.564T, and latent heat of fusion is 79.7 cal/gH2O.
Heat fusion is about 80 cal/g so your 79.7 cal/g (probably a better number than my memory) should be ok. However, I remember the heat vaporization as about 540 cal/g. But hold on ! You don't have heat vap anywhere in the equation.
You need specific heat ice
You need heat fusion water/ice.
You need specific heat liquid water.
Nowhere do you need heat vaporization for water. Somewhere you are substituting incorrectly. Allow me to use round numbers.
q1 = [120 x 0.500 x (8)] [Note: I think I remember specific heat ice = about 0.5 cal/g but check that out.]
q2 = [120 x 80]
q3 = [120 x 1 x (Tf-0)]
q4 = [500 x 1 x (Tf - 25)]
I think the answer is in the neighborhood of 15-20 C.
To find the final temperature of the water when all the ice has melted, you need to use the principle of conservation of energy.
First, let's calculate the energy required to melt the ice. The heat required to melt the ice can be calculated using the equation Q = m * L, where Q is the heat energy, m is the mass, and L is the latent heat of fusion.
The mass of ice is given as 120 g, and the latent heat of fusion of water is 334 J/g. So, the heat energy required to melt the ice can be calculated as:
Q = 120 g * 334 J/g = 40080 J
Next, let's calculate the heat lost by the water as it cools down to the final temperature. This can be determined using the equation Q = m * C * ΔT, where Q is the heat energy lost, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
The mass of water is given as 500 mL, which is equivalent to 500 g (since 1 mL of water is approximately equal to 1 g). The specific heat capacity of water is 4.18 J/g°C.
Assuming the final temperature of the water is T°C, the change in temperature can be calculated as (25 - T)°C.
So, the heat energy lost by the water can be calculated as:
Q = 500 g * 4.18 J/g°C * (25 - T)°C
According to the principle of conservation of energy, the heat energy lost by the water must equal the heat energy gained by the ice when it melts. Therefore, we can equate the two equations:
Q = 40080 J = 500 g * 4.18 J/g°C * (25 - T)°C
Now, we can solve for T:
40080 = 500 * 4.18 * (25 - T)
Divide both sides of the equation by (500 * 4.18):
(25 - T) = 40080 / (500 * 4.18)
(25 - T) = 15.326
Subtract 15.326 from both sides of the equation:
T = 25 - 15.326
T ≈ 9.7°C
Therefore, the final temperature of the water when all the ice has melted will be approximately 9.7°C.