What amount of ice must be added to 540.0 g of water at 25.0 °C to cool the water to 0.0 °C and have no ice
To determine the amount of ice needed to cool water from 25.0 °C to 0.0 °C without any remaining ice, we need to calculate the heat exchanged between the water and ice.
The equation for heat exchanged during a phase change is:
q = m * ΔHf
where:
q = heat exchanged (in Joules)
m = mass of the substance (in grams)
ΔHf = heat of fusion (in J/g)
The specific heat capacity of water is 4.18 J/(g·°C), which means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 °C.
Step 1: Determine the heat lost by the water.
The heat lost by the water can be calculated using the following formula:
q = m * c * ΔT
where:
q = heat lost (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (in J/(g·°C))
ΔT = change in temperature (in °C)
Given:
m = 540.0 g (mass of water)
c = 4.18 J/(g·°C) (specific heat capacity of water)
ΔT = 25.0 °C - 0.0 °C = 25.0 °C (change in temperature)
q = 540.0 g * 4.18 J/(g·°C) * 25.0 °C
q = 56535 J
Step 2: Determine the mass of ice needed.
The heat lost by the water will be gained by the ice as it undergoes a phase change from solid to liquid at 0.0 °C. The heat of fusion (ΔHf) for ice is 334 J/g.
Using the equation from step 1 and setting it equal to the heat exchanged during the phase change:
q = m * ΔHf
56535 J = m * 334 J/g
Solving for m:
m = 56535 J / 334 J/g
m ≈ 169.3 g
Therefore, approximately 169.3 grams of ice must be added to cool the 540.0 grams of water from 25.0 °C to 0.0 °C and have no remaining ice.
To determine the amount of ice needed to cool the water, we need to use the concept of heat transfer and the specific heat capacity equation.
The specific heat capacity, denoted as c, is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius per unit mass. The specific heat capacity for water is approximately 4.18 J/g°C.
The heat released or absorbed during a phase change is given by the equation: Q = m * ΔH, where Q is the heat energy, m is the mass, and ΔH is the enthalpy change.
In this case, the water will cool from 25.0 °C to 0.0 °C, meaning there will be a temperature change of 25.0 °C.
Using the equation Q = m * c * ΔT, we can calculate the heat energy released by the water: Q_water = m_water * c_water * ΔT_water.
Q_water = (540.0 g) * (4.18 J/g°C) * (25.0 °C)
Q_water = 56,115 J
To completely cool the water to 0.0 °C without any ice remaining, an equal amount of heat energy needs to be removed from the water and transferred to the ice.
Since the heat of fusion for water is 334 J/g, we can use the equation Q_ice = m_ice * ΔH_fusion to determine the mass of ice required.
Q_ice = m_ice * ΔH_fusion
56,115 J = m_ice * (334 J/g)
m_ice = 56,115 J / (334 J/g)
m_ice ≈ 167.87 g
Therefore, approximately 167.87 grams of ice must be added to the 540.0 grams of water at 25.0 °C in order to cool the water to 0.0 °C and have no ice remaining.
Well, I'm not quite sure about the exact amount of ice you'll need, but I can say that it's going to be "cool" to figure it out! Now, let's break it down:
To cool the water from 25.0 °C to 0.0 °C, you'll need to remove heat from the water. The heat lost by the water will be gained by the ice, causing it to melt and reach the same temperature as the water.
The heat lost by the water can be calculated using the formula:
Q = mcΔT
Where:
Q = heat lost by the water (in Joules)
m = mass of the water (in grams) = 540.0 g
c = specific heat capacity of water = 4.18 J/g°C (approximately)
ΔT = change in temperature = 25.0 °C - 0.0 °C = 25.0 °C
Now, the heat gained by the ice when it melts can be calculated using the formula:
Q = mLf
Where:
Q = heat gained by the ice (in Joules)
m = mass of the ice (in grams)
Lf = heat of fusion of ice = 334 J/g (approximately)
Since the heat lost by the water is equal to the heat gained by the ice, we can set the two equations equal to each other and solve for the mass of ice (m):
mcΔT = mLf
540.0 g * 4.18 J/g°C * 25.0 °C = m * 334 J/g
Solving for m, we find:
m ≈ (540.0 g * 4.18 J/g°C * 25.0 °C) / 334 J/g
m ≈ 3291.62 g
So, you'll need approximately 3291.62 grams of ice to cool the water to 0.0 °C and have no ice left. However, keep in mind that this is just an estimate and might not be "ice-act" depending on various factors like heat transfer efficiency and container insulation.
melting ice at zero is
mass ice*heat fusion = ?
Cooling 540 g water from 25 to zero is
mass H2O * specific heat ice x (Tfinal-Tinitial). Put those together to obtain
(mass ice*heat fusion) + (mass H2O x specific heat water x (Tfinal-Tinitial) = 0
Solve for mass ice.