How much heat energy is released when a 500g sample of water vapor is cooled from 120°C to a final
temperature of -20°C? [5T] [1C]
Specific Heat Capacity
𝑐𝑖𝑐𝑒 = 2.11 × 10^3
J/
kg°C
𝑐𝑤𝑎𝑡𝑒𝑟 = 2.11 × 10^3
J/
kg°C
𝑐𝑣𝑎𝑝𝑜𝑟 = 2.11 × 10^3
J/
kg°C
Latent Heat for Water
𝐿𝑓 = 3.4 × 10^5
J
/kg
𝐿𝑣 = 2.3 × 10^6
J
/kg
To find the amount of heat energy released, we need to calculate the heat released during the cooling process and the heat released during the phase change from vapor to liquid.
1) Heat released during cooling:
The formula to calculate the heat released during cooling is given by:
Q = mcΔT
Where:
Q = heat energy released
m = mass of the sample (500g)
c = specific heat capacity of the substance (2.11 × 10^3 J/kg°C)
ΔT = change in temperature (final temperature - initial temperature)
ΔT = (-20°C) - (120°C) = -140°C
Q = (500g) * (2.11 × 10^3 J/kg°C) * (-140°C)
Q = -1.47 × 10^8 J
Therefore, the heat released during cooling is -1.47 × 10^8 J.
2) Heat released during phase change:
The formula to calculate the heat released during a phase change is given by:
Q = mL
Where:
Q = heat energy released
m = mass of the sample (500g)
L = latent heat for the specific phase change (2.3 × 10^6 J/kg)
Q = (500g) * (2.3 × 10^6 J/kg)
Q = 1.15 × 10^9 J
Therefore, the heat released during the phase change is 1.15 × 10^9 J.
Adding the heat released during cooling and the heat released during the phase change:
Total heat energy released = -1.47 × 10^8 J + 1.15 × 10^9 J
Total heat energy released = 9.7 × 10^8 J
Therefore, the total amount of heat energy released when the 500g sample of water vapor is cooled from 120°C to a final temperature of -20°C is 9.7 × 10^8 J.