Water is sprayed on oranges during a frosty night. If an average of 11.8 g of water freezes on each orange how much heat is released.
Can you please help me set this problem up. I don know what formula to use?
thank you
q = mass H2O x heat fusion
You know mass H2O in grams, look up the heat fusion in J/g and solve for q in joules. The problem actually is not complete because it doesn't tell you the temperature of the water being sprayed. I think it assumes you will assume the temperature of the water when it hits the orange will be zero C (either that or that the water is cooled by the "frosty air" to zero and then freezes).
To calculate the heat released when water freezes, we need to use the formula for heat transfer.
The formula for heat transfer is:
Q = m * ΔHf
Where:
Q is the heat released (in joules)
m is the mass of the substance (in grams)
ΔHf is the heat of fusion (in J/g)
In this case, we need to find the heat released when water freezes on each orange. We are given the average mass of water frozen on each orange, which is 11.8 g.
The heat of fusion for water is 334 J/g.
Now, we can substitute the given values into the formula and calculate the heat released:
Q = 11.8 g * 334 J/g
Let's calculate the result:
Q = 3935.2 J
Therefore, approximately 3935.2 J of heat is released when 11.8 g of water freezes on each orange.
To determine the heat released when water freezes on oranges, you need to know the amount of water that freezes and the heat of fusion for water.
The formula that relates heat released, mass, and heat of fusion is:
Q = m * Hf
Where:
Q is the heat released (in Joules)
m is the mass of the water that freezes (in grams)
Hf is the heat of fusion for water (in Joules/gram)
In this case, the mass of water that freezes on each orange is given as 11.8 grams. However, the heat of fusion for water is not provided. The heat of fusion for water is the amount of heat energy required to change 1 gram of water from a liquid to a solid state at its freezing point (0 degrees Celsius).
The heat of fusion for water is approximately 334 J/g.
Using the provided mass of water that freezes on each orange (11.8 g) and the heat of fusion for water (334 J/g), you can now calculate the heat released.
Q = 11.8 g * 334 J/g
Now you can proceed to calculate the heat released by multiplying the mass of water by the heat of fusion for water.
Q = 3941 J
Therefore, approximately 3941 Joules of heat are released when an average of 11.8 g of water freezes on each orange.