Chapters 13 & 14. Temperature and Heat. Heat. Heat is a form of Energy Transfer Heat flows from areas of high energy to areas of lower energy Heat is transferred three way Conduction – requires contact Convection – mass movement of molecules Radiation – transfer over a distance.
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Temperature and Heat
C = 5/9 (F – 32)
Ex: C = 5/9 (212-32) = 100
F = (9/5 C) + 32
Ex: F = (9/5)(100) + 32 = 212
K = C +273
Ex: K = 100 + 273 = 373
C = K - 273
Ex: K = 373 - 273 = 100
Δl = (α)(l)(Δt)
l’ = l + Δl
Δl = (β)(V)(Δt)
V’ = V + ΔV
Δl = (α)(l)(Δt) = (19 x 10 -6)(0.03 m)(100 °C)=
l’ = Δl + l = 0.000057 m + 0.03 m =
Q = mCpΔt
(5 kg)(Cp = 4186 J/kg°C)(100°C)
= 493000 Joules
Q copper = Q water
mC copperΔt= mC water Δt
(5 kg)(390 J/Kg°C)(70°C)=m(4186 J/Kg°C)(10°C)
m = 3.02 kg
Both of these changes happen at the same point; 0° for water!
Vapor is the gas phase of any substance which is normally a liquid at room temperature!
Q = mΔHf for Fusion
Q = mΔHv for Vaporization
How much heat is required to melt 5 kg of ice at 0° C?
Q = mΔHf = (5 kg)(3.33 x 105 J/kg °C) =
1.665 x 106 J
Ice - Water
How much energy is required to convert 5 kg of ice at -5 °C to Steam at 120 °C?
Raise temperature of ice to melting point
Raise temperature of Water to boiling point
Raise Temperature of Steam to 120 °C
Step 1: Raise temperature of Ice
Q = mCΔt = (5 Kg)(2100 J/kg°C)(5 °C)= 52500 J
Step 2: Melt Ice
Q = mΔHf = (5 Kg)(3.33 x 105 J/kg) = 1665000 J
Step 3: Raise temperature of water to boiling
Q = mCΔt = (5 Kg)(4186 J/kg°C)(100 °C)= 2093000 J
Step 4: Vaporize Water
Q = mΔHv = (5 Kg)(22.6 x 105 J/kg) = 11300000 J
Step 5: Raise temperature of Steam to 120°C
Q = mCΔt = (5 Kg)(2010 J/kg°C)(20 °C)= 201000 J
Step 6: Get Total
Qtotal= 52500 + 1665000 + 2093000 + 11300000 + 201000 = 15311500J