The enthalpy of fusion of ice is 6.020 kJ/mol. The heat capacity of liquid water is 75.4 J/mol degrees Celsius. What is the smallest number of ice cubes at 0 degrees Celsius, each containing one mole of water, necessary to cool 500. g of liquid water initially at 20 degrees Celsius to 0 degrees Celsius?
very good question and it is very nicely answered and thanks for the answer
Why did the ice cubes go to therapy? Because they were feeling a bit watered down! Okay, let's solve this freezing puzzle for you.
To cool down the liquid water from 20 degrees Celsius to 0 degrees Celsius, we need to calculate the amount of heat that needs to be removed. The heat capacity of liquid water tells us that it takes 75.4 J to raise the temperature of 1 mole of water by 1 degree Celsius.
The temperature change is 20 degrees Celsius, so to cool down the water, we need to remove:
Q = heat capacity × moles × temperature change
Q = 75.4 J/mol°C × 1 mol × 20°C
Q = 1508 J
Now, let's calculate how many ice cubes we need. The enthalpy of fusion of ice tells us that 6.020 kJ are released or absorbed when 1 mole of ice melts or freezes, respectively.
We want to cool down 500 g of water, which is equivalent to 0.5 kg or 0.5 moles of water.
The enthalpy change can be calculated as:
Enthalpy Change = enthalpy of fusion × moles
Enthalpy Change = 6.020 kJ/mol × 0.5 mol
Enthalpy Change = 3.010 kJ
So, to cool down 500 g of liquid water from 20°C to 0°C, we need to remove 1508 J of heat, which is equivalent to 1.508 kJ. However, each mole of ice releases 3.010 kJ upon freezing. Therefore, the smallest number of ice cubes needed would be:
Number of Ice Cubes = (Total Energy Removed) / (Enthalpy Change per Ice Cube)
Number of Ice Cubes = 1.508 kJ / 3.010 kJ
Number of Ice Cubes ≈ 0.5 ice cubes
Now, I know we can't have half an ice cube, so in this case, you would need at least 1 ice cube to cool down the liquid water from 20°C to 0°C. Remember to always keep your cool, just like those ice cubes!
To determine the number of ice cubes needed to cool the liquid water to 0 degrees Celsius, we need to calculate the heat released by the liquid water as it cools down.
First, let's calculate the heat released when the liquid water cools from 20 degrees Celsius to 0 degrees Celsius.
The heat capacity of water is given as 75.4 J/mol degrees Celsius. Since we have 500 g of liquid water, we need to convert it to moles.
The molar mass of water is approximately 18.015 g/mol.
Number of moles of water present in 500 g = (500 g) / (18.015 g/mol) = 27.74 mol
Now, let's calculate the heat released when the liquid water cools from 20 degrees Celsius to 0 degrees Celsius.
ΔT = 20 degrees Celsius - 0 degrees Celsius = 20 degrees Celsius
Q = heat released = (heat capacity of water) * (number of moles of water) * (ΔT)
= (75.4 J/mol degrees Celsius) * (27.74 mol) * (20 degrees Celsius)
= 41,134 J
Next, let's calculate the heat required to melt the ice cubes at 0 degrees Celsius.
The enthalpy of fusion of ice is given as 6.020 kJ/mol. Since we know that one mole of water is present in each ice cube, the heat required to melt the ice cubes can now be calculated.
Heat required to melt the ice = (enthalpy of fusion) * (number of ice cubes)
= (6.020 kJ/mol) * (number of ice cubes)
To calculate the number of ice cubes, we need to convert the heat released when the liquid water cools to Joules.
Heat released = 41,134 J = 41.134 kJ
Setting the heat released equal to the heat required to melt the ice cubes:
41.134 kJ = (6.020 kJ/mol) * (number of ice cubes)
Solving for the number of ice cubes:
(number of ice cubes) = 41.134 kJ / 6.020 kJ/mol ≈ 6.84 mol
Since each ice cube contains 1 mole of water, we need to round up to the nearest whole number.
Therefore, the smallest number of ice cubes required is 7.
To find the smallest number of ice cubes required to cool the liquid water from 20 degrees Celsius to 0 degrees Celsius, we need to calculate the heat transfer for each step.
First, let's calculate the amount of heat required to cool the liquid water from 20 degrees Celsius to 0 degrees Celsius.
1. Calculate the heat transferred to the water using the formula:
Q = m * C * ΔT
where Q is the heat transferred, m is the mass of water, C is the heat capacity of water, and ΔT is the change in temperature.
Given:
m = 500 g = 0.5 kg (since 1 kg = 1000 g)
C = 75.4 J/mol degrees Celsius
ΔT = 20 degrees Celsius - 0 degrees Celsius = 20 degrees Celsius
Plugging in the values:
Q = 0.5 kg * 75.4 J/mol degrees Celsius * 20 degrees Celsius
Q = 753 J
So, the amount of heat required to cool the liquid water from 20 degrees Celsius to 0 degrees Celsius is 753 J.
Next, let's calculate the amount of heat released when the ice cubes at 0 degrees Celsius turn into liquid water at 0 degrees Celsius.
2. Calculate the heat released during the phase change, which is the enthalpy of fusion, using the formula:
Q = n * ΔH_fusion
where Q is the heat released, n is the number of moles, and ΔH_fusion is the enthalpy of fusion.
Given:
ΔH_fusion = 6.020 kJ/mol = 6.020 * 10^3 J/mol (since 1 kJ = 10^3 J)
Since the number of moles of water is 1 mole per ice cube, the heat released when one ice cube melts is 6.020 * 10^3 J.
To find the smallest number of ice cubes required to cool the water, divide the total heat transferred (753 J) by the heat released per ice cube (6.020 * 10^3 J) and round up to the nearest whole number.
Number of ice cubes = ceil(Q / (ΔH_fusion / n))
Using the values calculated above:
Number of ice cubes = ceil(753 J / (6.020 * 10^3 J)) = ceil(0.125)
Therefore, the smallest number of ice cubes needed is 1.
sum of heats gained=0
heat gained by ice+heatgained by water=0
N*6020 + (500/18)*75.4*(0-20)=0
solve for N, take the next highest whole number.