Calculate the heat required to convert 5gm of ice at 0oc into 5gm of steam at 100oc.
q1 = heat to melt ice at zero C to liquid water at zero C.
q1 = mass ice x heat fusion = ?
q2 = heat to change T liquid H2O at zero C to liquid H2O @ 100 C.
q2 = mass water x specific heat H2O x (Tfinal-Tinitial) = ?
q3 = heat to change liquid H2O at 100 C to steam at 100 C.
q3 = mass H2O x heat vaporization = ?
total q = q1 + q2 + q3 = ?
I got 3600 cals = 15,062 joules = 15.1 Kj
To calculate the heat required to convert a substance from one state to another, you can use the formula:
Q = m * L
where:
Q is the heat energy required in Joules (J),
m is the mass of the substance in grams (g), and
L is the specific heat of the substance in J/g
For the given problem, we need to consider two different processes:
1. Heating the ice from 0°C to its melting point
2. Converting the liquid water at 0°C to steam at 100°C
Let's break down the calculation into steps:
Step 1: Heating the ice to its melting point
The specific heat of ice is 2.09 J/g°C.
Since the temperature change is from 0°C to 0°C, there is no change in temperature. However, we still need to provide heat energy to reach the melting point.
So, the heat required to heat the ice to its melting point would be:
Q1 = m * L1
= 5g * 2.09 J/g°C * 0°C
= 0 J (as there is no change in temperature)
Step 2: Melting the ice
To convert the ice at 0°C to liquid water at 0°C, we need to provide the latent heat of fusion.
The latent heat of fusion for water is 334 J/g.
The heat required to melt the ice would be:
Q2 = m * L2
= 5g * 334 J/g
= 1670 J
Step 3: Heating the water to boiling point (100°C)
The specific heat of water is 4.18 J/g°C.
Since the temperature change is from 0°C to 100°C, we need to calculate the heat required for this temperature range.
The heat required to heat the water to boiling point would be:
Q3 = m * L3 * ΔT
= 5g * 4.18 J/g°C * (100°C - 0°C)
= 2090 J
Step 4: Vaporizing the water
To convert the water at 100°C to steam at 100°C, we need to provide the latent heat of vaporization.
The latent heat of vaporization for water is 2260 J/g.
The heat required to vaporize the water would be:
Q4 = m * L4
= 5g * 2260 J/g
= 11300 J
Step 5: Heating the steam to the final temperature (100°C)
Since the steam is already at its boiling point, there is no further temperature change, but we still need to provide heat energy.
The heat required to heat the steam to its final temperature would be:
Q5 = m * L5
= 5g * 2.09 J/g°C * 0°C
= 0 J (as there is no change in temperature)
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 0 J + 1670 J + 2090 J + 11300 J + 0 J
= 15060 J or 15.06 kJ
Therefore, to convert 5 grams of ice at 0°C into 5 grams of steam at 100°C, you need a total of 15.06 kilojoules (kJ) of heat.