How much heat energy, in kilojoules, is required to convert 48.0g of ice at −18.0 ∘C to water at 25.0 ∘C ?
Please show all steps. i don't know where to start
There are three equations that handle all of these situations.
Equation 1 is for WITHIN a phase,
q = mass x specific heat x (Tfinal-Tinitial).
Example. For the liquid phase of water from zero C to 25 C, then q for that part is
q = (48g x 4.184 J/g*C x (25-0) = ?
Equations 2 and 3 are at a phase change (for water that is melting ice at zero C or boiling at 100 C to steam).
q = mass x heat fusion for melting or
q = mass x heat vaporization for boiling.
Example. For melting ice at zero to liquid water (still at zero)
q = 48g ice x 334 J/g.
So you have to move ice at -18.0 C to 25.0 C and you do it in steps.
q1 = heat to raise T of ice at -18.0C to ice at zero C. That's equation 1.
q2 = heat to melt ice at zero to liquid at zero. That's equation 2.
q3 = heat to raise T of liquid water from zero C to 25 C. That's equation 1.
Total q is q1 + q2 + q3.
To determine the amount of heat energy required to convert ice to water, several steps need to be taken:
1. Calculate the heat energy required to raise the temperature of the ice from -18.0 °C to its melting point, 0 °C.
2. Determine the heat energy required to melt the ice at 0 °C.
3. Calculate the heat energy required to raise the temperature of the melted ice to 25.0 °C.
Let's begin:
Step 1: Calculating heat energy to raise the temperature of the ice from -18.0 °C to 0 °C.
The specific heat capacity of ice is 2.09 J/g·°C.
The formula used to calculate the heat energy is:
q1 = m × c × ΔT
Where:
q1 = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
Given that:
m = 48.0 g (mass)
c = 2.09 J/g·°C (specific heat capacity)
ΔT = 0 °C - (-18.0 °C) = 18.0 °C
Plugging the values into the formula:
q1 = 48.0 g × 2.09 J/g·°C × 18.0 °C = 1804.32 J
Since the answer requested is in kilojoules, divide the result by 1000:
q1 = 1804.32 J / 1000 = 1.80432 kJ
Step 2: Calculating heat energy to melt the ice at 0 °C.
The heat of fusion for water is 334 J/g.
The formula used to calculate the heat energy is:
q2 = m × ΔHf
Where:
q2 = heat energy
m = mass
ΔHf = heat of fusion
Given that:
m = 48.0 g (mass)
ΔHf = 334 J/g (heat of fusion)
Plugging the values into the formula:
q2 = 48.0 g × 334 J/g = 16032 J
Again, divide the result by 1000 to convert to kilojoules:
q2 = 16032 J / 1000 = 16.032 kJ
Step 3: Calculating heat energy to raise the temperature of the melted ice to 25.0 °C.
The specific heat capacity of water is 4.18 J/g·°C.
Using the same formula as in step 1:
q3 = m × c × ΔT
Given that:
m = 48.0 g (mass)
c = 4.18 J/g·°C (specific heat capacity)
ΔT = 25.0 °C - 0 °C = 25.0 °C
Plugging the values into the formula:
q3 = 48.0 g × 4.18 J/g·°C × 25.0 °C = 4992.6 J
Divide the result by 1000 to convert to kilojoules:
q3 = 4992.6 J / 1000 = 4.9926 kJ
Finally, add the three amounts of heat energy together to find the total heat energy required:
Total heat energy = q1 + q2 + q3
Total heat energy = 1.80432 kJ + 16.032 kJ + 4.9926 kJ
Calculating the sum:
Total heat energy = 22.82892 kJ
Therefore, approximately 22.8 kJ of heat energy is required to convert 48.0 g of ice at -18.0 °C to water at 25.0 °C.
To determine the amount of heat energy required to convert ice at -18.0 °C to water at 25.0 °C, we need to consider two separate processes:
1. Heating the ice from -18.0 °C to 0 °C (melting process).
2. Heating the ice-water mixture from 0 °C to 25.0 °C (heating process).
Let's break down the steps to calculate the heat energy required for each process:
1. Calculating the heat energy required to heat the ice from -18.0 °C to 0 °C (melting process):
The specific heat capacity of ice (c_ice) is 2.09 J/g·°C.
The heat equation for this step is Q = m × c × ΔT, where
- Q is the heat energy,
- m is the mass,
- c is the specific heat capacity, and
- ΔT is the change in temperature.
Substituting the given values into the equation:
Q_1 = m × c_ice × ΔT_1
= 48.0 g × 2.09 J/g·°C × (0 °C - (-18.0 °C))
= 48.0 g × 2.09 J/g·°C × 18.0 °C
Now, let's convert this answer to kilojoules (kJ):
Q_1 = (48.0 g × 2.09 J/g·°C × 18.0 °C) ÷ 1000
≈ 1.79 kJ
2. Calculating the heat energy required to heat the ice-water mixture from 0 °C to 25.0 °C (heating process):
The specific heat capacity of water (c_water) is 4.18 J/g·°C.
The heat equation for this step is Q = m × c × ΔT, where
- Q is the heat energy,
- m is the mass,
- c is the specific heat capacity, and
- ΔT is the change in temperature.
Substituting the given values into the equation:
Q_2 = m × c_water × ΔT_2
= 48.0 g × 4.18 J/g·°C × (25.0 °C - 0 °C)
= 48.0 g × 4.18 J/g·°C × 25.0 °C
Again, let's convert this answer to kilojoules (kJ):
Q_2 = (48.0 g × 4.18 J/g·°C × 25.0 °C) ÷ 1000
≈ 49.9 kJ
Finally, to get the total heat energy required, we add up the values of Q_1 and Q_2:
Total heat energy = Q_1 + Q_2
= 1.79 kJ + 49.9 kJ
≈ 51.7 kJ
Therefore, approximately 51.7 kilojoules (kJ) of heat energy are required to convert 48.0 grams of ice at -18.0 °C to water at 25.0 °C.