What is the heat required in Joules to raise the temperature of 100.0g ice at
-20 degrees C to steam at 120 degrees C?
To calculate the heat required, you need to consider the different stages of heating and cooling. The process can be broken down into three steps:
1. Heating the ice from -20 degrees Celsius to 0 degrees Celsius (solid phase).
2. Melting the ice at 0 degrees Celsius (phase change).
3. Heating the liquid water from 0 degrees Celsius to 100 degrees Celsius (liquid phase).
4. Boiling the water at 100 degrees Celsius (phase change).
5. Heating the steam from 100 degrees Celsius to 120 degrees Celsius (gaseous phase).
To calculate the heat required for each step, you need to use the formulas:
Q = m × c × ΔT (for heating/cooling, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature)
Q = m × L (for phase changes, where L is the specific latent heat)
Step 1: Heating the ice from -20 degrees Celsius to 0 degrees Celsius:
The specific heat capacity of ice is 2.09 J/g°C.
ΔT = 0°C - (-20°C) = 20°C
Q1 = 100.0g × 2.09 J/g°C × 20°C = 4180 J
Step 2: Melting the ice at 0 degrees Celsius:
The specific latent heat of fusion for ice is 334 J/g.
Q2 = 100.0g × 334 J/g = 33400 J
Step 3: Heating the liquid water from 0 degrees Celsius to 100 degrees Celsius:
The specific heat capacity of water is 4.18 J/g°C.
ΔT = 100°C - 0°C = 100°C
Q3 = 100.0g × 4.18 J/g°C × 100°C = 41800 J
Step 4: Boiling the water at 100 degrees Celsius:
The specific latent heat of vaporization for water is 2260 J/g.
Q4 = 100.0g × 2260 J/g = 226000 J
Step 5: Heating the steam from 100 degrees Celsius to 120 degrees Celsius:
The specific heat capacity of steam is 2.03 J/g°C.
ΔT = 120°C - 100°C = 20°C
Q5 = 100.0g × 2.03 J/g°C × 20°C = 4060 J
Finally, add up all the individual heat values to find the total heat required:
Total Q = Q1 + Q2 + Q3 + Q4 + Q5 = 4180 J + 33400 J + 41800 J + 226000 J + 4060 J = 311440 J
Therefore, the heat required in Joules to raise the temperature of 100.0g of ice at -20 degrees Celsius to steam at 120 degrees Celsius is 311440 J.
to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed
now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from ice->water->steam
thus we need another data called it Latent Heat of Fusion and Latent Heat of Vaporization to calculate for the heat required to change its phase:
H1 = m(Lf)
H2 = m(Lv)
where
m = mass
Lf = Latent Heat of Fusion (fusion means melting)
Lv = Latent Heat of Vaporization (vaporization means from liquid, it becomes vapor)
thus for the problem,
H1 = 100*Lf
H2 = 100*Lv
also, the heat required to raise the ICE's temp (from -20 C to 0 C) is
Q1 = mc(0 -(-20)) = 100*20*c = 100*20*c
the heat required to raise the LIQUID's temp (from 0 C to 100 C) is
Q2 = mc(100 - 0) = 100*75*c = 100*100*c
the heat required to raise the STEAM's temp (from 100 to 135) is
Q3 = mc(120 - 100) = 100*20*c
thus the total heat needed is:
Q,total = Q1 + H1 + Q2 + H2 + Q3
note that can find c, Lf and Lv of water in books or you can just google them. remember and be careful of the units.
hope this helps~ :)