How many joules are required to convert 16.5 g of ice at 0.0°C to liquid water at 32.0°C? The heat of fusion of water is 334 J/g.
To calculate the amount of energy required to convert the ice to water, we will need to consider two steps:
Step 1: Calculate the energy required to raise the temperature of the ice from 0.0°C to its melting point of 0.0°C.
Step 2: Calculate the energy required to melt the ice at 0.0°C to liquid water at 0.0°C.
Step 3: Calculate the energy required to raise the temperature of the liquid water from 0.0°C to 32.0°C.
Step 1: Calculate the energy to raise the temperature of the ice from 0.0°C to its melting point of 0.0°C.
The specific heat capacity of ice is 2.09 J/g·°C.
Energy = mass × specific heat capacity × temperature change
Energy = 16.5 g × 2.09 J/g·°C × (0-0.0)°C
Energy = 0 J
Since there is no temperature change, no energy is required in this step.
Step 2: Calculate the energy required to melt the ice at 0.0°C to liquid water at 0.0°C.
The heat of fusion of water is 334 J/g.
Energy = mass × heat of fusion
Energy = 16.5 g × 334 J/g
Energy = 5511 J
Step 3: Calculate the energy required to raise the temperature of the liquid water from 0.0°C to 32.0°C.
The specific heat capacity of water is 4.18 J/g·°C.
Energy = mass × specific heat capacity × temperature change
Energy = 16.5 g × 4.18 J/g·°C × (32.0-0.0)°C
Energy = 2187.12 J
Finally, to find the total energy required, add the energy from each step.
Total Energy = Energy from step 2 + Energy from step 3
Total Energy = 5511 J + 2187.12 J
Total Energy ≈ 7698.12 J (rounded to two decimal places)
Therefore, approximately 7698.12 joules of energy are required to convert 16.5 g of ice at 0.0°C to liquid water at 32.0°C using the given heat of fusion.
To find the total amount of energy required to convert ice to liquid water, we need to consider two steps:
Step 1: Heating the ice from its initial temperature to its melting point (0.0°C).
Step 2: Melting the ice at 0.0°C to liquid water at the final temperature (32.0°C).
Step 1: Calculating the energy required to heat the ice to its melting point:
We can use the specific heat capacity equation:
q1 = m * C * ΔT
where:
q1 = energy (in joules)
m = mass of the ice (in grams)
C = specific heat capacity of ice (in J/g°C)
ΔT = change in temperature
In this case, since the ice is at 0.0°C and we want to heat it to 0.0°C, ΔT = 0.0°C - 0.0°C = 0.0°C.
Substituting the values given:
m = 16.5 g
C = specific heat capacity of ice = 2.09 J/g°C (approximate value for ice)
ΔT = 0.0°C - 0.0°C = 0.0°C
q1 = 16.5 g * 2.09 J/g°C * 0.0°C
q1 = 0 Joules
Therefore, no additional energy is required to heat the ice to its melting point since it is already at 0.0°C.
Step 2: Calculating the energy required to melt the ice:
We can use the equation:
q2 = m * ΔHf
where:
q2 = energy (in joules)
m = mass of the ice (in grams)
ΔHf = heat of fusion of water (in J/g)
Substituting the values given:
m = 16.5 g
ΔHf = 334 J/g
q2 = 16.5 g * 334 J/g
q2 = 5501 Joules
Therefore, the energy required to convert 16.5 g of ice to liquid water is 5501 Joules.
add the heats involved
heat to melt ice at 0C
heat to heat water to 32C
16.5*Hf
16.5*specifheatwater*(32.0-0)