A 55.0-g piece of ice at 0.0 is added to a sample of water at 8.0. All of the ice melts and the temperature of the water decreases to 0.0.
How many kg of water were in the sample?
(mass ice x heat fusion ice) + [mass melted ice x specific heat melted ice x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for mass H2O (third term) and convert to kg.
To determine the mass of water in the sample, we need to use the principle of conservation of energy.
First, let's calculate the energy required to melt the ice. The amount of energy required to melt a substance can be calculated using the formula:
Q = m * ΔHf
where Q is the energy, m is the mass of the substance, and ΔHf is the heat of fusion, which is the amount of energy required to convert one gram of a substance from solid to liquid at its melting point.
For ice, the heat of fusion is 334 J/g.
Given:
Mass of ice = 55.0 g
Change in temperature for ice = 0.0 °C
Using the formula, we can calculate the energy required to melt the ice:
Q_ice = (55.0 g) * (334 J/g)
Q_ice = 18370 J
Since the ice melts at 0.0 °C, it requires energy without a change in temperature.
Next, let's calculate the energy lost by the water to cool down from 8.0 °C to 0.0 °C. The specific heat capacity of water is approximately 4.18 J/g°C.
Given:
Change in temperature for water = 8.0 °C - 0.0 °C = 8.0 °C
Using the formula for energy:
Q_water = m * c * ΔT
where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q_water = m * c * ΔT
Q_water = m * (4.18 J/g°C) * (8.0 °C)
Q_water = 33.44 m J
Now, since the energy lost by the water is equal to the energy gained by the ice (according to the principle of conservation of energy), we can set up an equation:
Q_ice = Q_water
18370 J = 33.44 m J
Solving for m, the mass of water:
m = 18370 J / 33.44 J/g
m ≈ 549.52 g
Therefore, there were approximately 549.52 grams of water in the sample. To convert this to kilograms, divide by 1000:
m ≈ 549.52 g / 1000
m ≈ 0.54952 kg
So, there were approximately 0.54952 kg of water in the sample.