Chapter five:
Thermodynamics
Intended Learning Outcomes:
In this chapter we will
• Distinguish between the heat and the temperature • Explain the specific heat capacity
• Solve problems using the heat equation • Explain the latent heat of fusion and the latent heat of evaporation. • Explore gas laws & the ideal gas equation
• Solve mathematical problems on the gas Laws • Explain 1 st law of thermodynamics
• Solve mathematical problems on the 1 st law of thermodynamics
5.1Thermal physics
5.1.1 Temperature and Heat
• Temperature: is a measure of hotness or coldness of an object (describes the average
kinetic energy of molecules within a material or system)
• Heat: the energy transferred between objects because of a temperature difference.
• Temperature Scales: Fahrenheit (T F ), Celsius (T C ), and Kelvin (T K )
• Here are the temperature conversion formulas:
Celsius to Kelvin: T K = T C + 273
Kelvin to Celsius: T C = T K - 273
Fahrenheit to Celsius: T C = (T F -32) (5/9)
Celsius to Fahrenheit: T F = T C (9/5) + 32
Fahrenheit to Kelvin: T K = (T F -32) (5/9) + 273
Kelvin to Fahrenheit: T F = (T K -273) (9/5) + 32
𝑇 𝑐 − 0/100 = 𝑇 𝑘 − 273/100= 𝑇 𝐹 − 32/180
Example 1 Convert 300 K to Celsius.
𝑇 𝑐 = 300 − 273 = 27°
Example 2
45°C is how many degrees Fahrenheit?
𝑇 𝑐 − 0/100 = 𝑇 𝐹 − 32/180
𝑇 𝐹 = 113 ℉
5.1.2 Specific Heat Capacity (S)
• The amount of heat (Q) required to increase the temperature of a mass m from T 1 to
T 2 is given by:
Where:
𝑄 = 𝑚 × 𝑠 × ∆𝑇
Q : Is the amount of heat (Joules) m : Is the mass (kg)
ΔT : Is the change in temperature, ΔT = T 2 - T 1 (K) S : Is constant called the Specific heat capacity
• Specific heat capacity (S): the amount of heat needed to raise the temperature of 1 kg
of substances by 1 o C. ( J/ kg. o C).
Example: 3
What does it mean by the specific heat capacity of aluminum is 900 J/kg °C?
Answer:
900 J of heat needs to be supplied to 1 kg of aluminum to increase its temperature by 1°C.
Example 4
Water has a specific heat capacity of 4200 J/kg °C. Calculate the energy required to raise the temperature of 0.4 kg of water from 30°C to 85°C.
Solution:
𝑄 = 𝑚 × 𝑠 × ∆𝑇
𝑄 = 0.4 × 4200 × (85 − 30) = 92400 𝐽
Example 5
Solution:
𝑄 = 𝑚 × 𝑠 × ∆𝑇
𝑄 = 0.4 × 4200 × (85 − 30) = 92400 𝐽
----------------------------------------------------------------------------------------------------------------A 2.5kg lump of copper is heated from 20°C to 120°C. The energy supplied by the heater is 105000 J. Calculate the specific heat capacity of the copper.
𝑄 = 𝑚 × 𝑠 × ∆𝑇
105000 = 2.5 × 𝑠 × (120 − 20)
s = 105000/ 2.5 × (120 − 20) = 420 𝐽/𝑘𝑔 °𝐶
Example 6 If 0.75 kg of water ( s = 4200 J/kg . 0 C) at a temperature of 10°C is heated using 16380 J of heat energy. Calculate the final temperature of the water.
Solution:
𝑄 = 𝑚 × 𝑠 × ∆𝑇 𝑄 = 𝑚 × 𝑠 × (𝑇 2 − 𝑇 1 ) 16380 = 0.75 × 4200 × (𝑇 2 − 10)
(𝑇 2 − 10) = 16380 /0.75 × 4200
(𝑇 2 ) = 15.2 ℃
5.1.3 Latent Heat of Fusion (L f ) ** (Additional part)
• Latent heat of fusion of a solid (L f ) is the amount of heat required to convert 1 kg of
solid material to 1 kg of liquid without change in temperature (at melting point temperature).
• 𝑄 = 𝑚 × 𝐿 𝑓
The unit of 𝐿 𝑓 is J/kg
5.1.4 Latent Heat of Vaporization (L v ) ** (Additional part)
• Latent heat of vaporization (L v ) of a liquid is the amount of heat required to convert
1 kg of liquid material to 1 kg of gas without change in temperature (at the boiling point temperature).
• 𝑄 = 𝑚 × 𝐿 𝑣
• The unit of 𝐿 𝑣 is J/kg
Example 7
4400 J of heat was supplied to 2kg of water at 100°C. All of the water turned into steam at 100°C. What is the specific latent heat of vaporization of water?
Solution:
𝑄 = 𝑚 × 𝐿 𝑣
4400 = 2 × 𝐿 𝑣
𝐿 𝑣 = 2200 J/kg
#General-physics
#generalphysics
#BMC
#كلية_البترجي
#كلية_البترجي_الطبية
12 окт 2024
