A quantity of ice at 0.0°C was added to 25.6 g of water at 21.0°C to give water at 0.0°C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g · °C).
I don't understand how to set this problem up (what the formula is and where to plug in which qualities). Any help would be very much appreciated.
How much heat must be removed from the 25.6 g H2O @ 21.0.0 C to move it to 0.0 C.
That will be
q = heat to be removed = mass x specific heat water x delta T.
q = 25.6 g x 4.18 x 21.0 = about 2000 J (but you need to do it exactly).
Then 2000 J x 1 g/heat fusion = grams ice.
You are given the heat of fusion in kJ/mol. You must change that to J/g so it will fit the remainder of the problem. It's approximately 334 J/g but you need to verify that. Post your work if you get stuck.
it worked perfectly, thank you so much
To solve this problem, we can use the principles of energy conservation and the formula for heat transfer.
The heat gained or lost by a substance can be calculated using the formula:
q = m * c * ΔT
where q is the heat gained or lost (in joules), m is the mass of the substance (in grams), c is the specific heat of the substance (in J/(g · °C)), and ΔT is the change in temperature of the substance (in °C).
In this problem, we need to find the mass of ice added to water. We can assume that the heat gained by the ice as it melts is equal to the heat lost by the water as it cools down.
The equation for the heat lost by water can be written as:
q_water = m_water * c_water * ΔT_water
The equation for the heat gained by ice can be written as:
q_ice = m_ice * ΔH_fusion
The heat gained by the ice is used to melt the ice, so it can be written as the heat required for the fusion of the ice. ΔH_fusion is the heat of fusion of water, which is 6.01 kJ/mol.
Since the water and the ice reach the same final temperature, we can set the heat gained by the ice equal to the heat lost by the water:
m_water * c_water * ΔT_water = m_ice * ΔH_fusion
We can rearrange this equation to solve for the mass of ice (m_ice):
m_ice = (m_water * c_water * ΔT_water) / ΔH_fusion
Plug in the given values:
m_water = 25.6 g
c_water = 4.18 J/(g · °C)
ΔT_water = (0.0°C - 21.0°C) = -21.0°C
ΔH_fusion = 6.01 kJ/mol = 6.01 * 10^3 J / (mol * °C) (Note: mol is not given, but it cancels out when solving for the mass of ice.)
Now substitute these values into the equation:
m_ice = (25.6 g * 4.18 J/(g · °C) * -21.0°C) / (6.01 * 10^3 J / (mol * °C))
Calculating this expression will give you the mass of ice added to the water, which will answer the question.