THERMAL EQUILIBRIUM

DE LA SALLE SANTIAGO ZOBEL SCHOOLHIGH SCHOOL SCIENCE DEPARTMENT

SY 2012-2013 / TERM 2PHYSICAL SCIENCE

MODULE 2.2 – HEAT AND TEMPERATURE

Lesson Objectives:

Recall how heat transfer happens;

Define thermal equilibrium;

Explain how the law of conservation of energy applies to heat transfer,; and objects in thermal equilibrium; andSystematically solve problems on calorimetry and methods of mixtures by applying the formula of the law of conservation of energy in thermal systems.

Objects in “Thermal Contact”If you mix cold milk with hot coffee,

Will heat transfer happen within the milk-coffee mixture? Why?What will be the direction of heat flow? Why?

After some time, what do you think will happen to the temperature of the milk-coffee mixture?

Thermal Equilibriumthe condition when objects in thermal contact achieve a common final temperaturein this state, no NET heat transfer happens between the two objects in thermal contactthe sum of all heat gained and heat lost in the thermal system is equal to zero

Object A Object B

The Law of Conservation of Energy

“Energy can neither be created nor destroyed.”

In a thermal system, this law applies such that no heat (energy) is created nor destroyed.Therefore, all heat gained should balance all heat lost in a thermal system.

𝑎 𝑙𝑙h𝑒𝑎𝑡𝑔𝑎𝑖𝑛𝑒𝑑=𝑎𝑙𝑙h𝑒𝑎𝑡 𝑙𝑜𝑠𝑡

Σ (+𝑄 )=Σ (−𝑄 )

𝑄1+𝑄2+𝑄3+ .. .𝑄𝑛=0

or

or

Objects in Thermal Equilibrium

When objects achieve the state of thermal equilibrium, the formula of the law of conservation of energy in thermal systems can be used to solve for the common final temperature of the system and other involved variables.Note that the formula𝑄1+𝑄2+𝑄3=0is just the same as 𝑚1𝑐1 (𝑇 𝑓 −𝑇 𝑖1 )+𝑚2𝑐2 (𝑇 𝑓 −𝑇 𝑖2 )+𝑚3𝑐3 (𝑇 𝑓 −𝑇 𝑖3 )=0

Sample ProblemA 100 oC nickel coin with a mass of 10-g is immersed in a 50-g glass of water at 20 oC. The specific heat of nickel is 0.44 J/g oC while water’s specific heat is 4.18 J/g oC. What is the final temperature of the system when it achieves thermal equilibrium?𝑆𝑡𝑒𝑝 1:𝑇𝑎𝑏𝑢𝑙𝑎𝑡𝑒 h𝑡 𝑒𝐺𝑖𝑣𝑒𝑛

𝑁𝑖𝑐𝑘𝑒𝑙𝐶𝑜𝑖𝑛 𝑊𝑎𝑡𝑒𝑟10 g0.44  J / g   oC100 oC

50 g4.18  J / g   oC20  oC

Required  To  Find   = ?

𝑄1+𝑄2=0

𝑚1𝑐1 (𝑇 𝑓 −𝑇 𝑖1 )+𝑚2𝑐2 (𝑇 𝑓 −𝑇 𝑖2 )=0

𝑚1𝑐1𝑇 𝑓 −𝑚1𝑐1𝑇 𝑖 1+𝑚2𝑐2𝑇 𝑓 −𝑚2𝑐2𝑇 𝑖2=0

𝑚1𝑐1𝑇 𝑓+𝑚2𝑐2𝑇 𝑓=𝑚1𝑐1𝑇 𝑖1+𝑚2𝑐2𝑇 𝑖 2

𝑇 𝑓 (𝑚¿¿1𝑐1+𝑚2𝑐2)=𝑚1𝑐1𝑇 𝑖 1+𝑚2𝑐2𝑇 𝑖 2¿𝑚1𝑐1+𝑚2𝑐2 𝑚1𝑐1+𝑚2𝑐2

𝑇 𝑓=𝑚1𝑐1𝑇 𝑖1+𝑚2𝑐2𝑇 𝑖2

𝑚1𝑐1+𝑚2𝑐2

𝑇 𝑓=(10𝑔)(0.44 𝐽

𝑔°𝐶 ) (100 °𝐶 )+ (50𝑔 )(4.18 𝐽𝑔°𝐶 ) (20 °𝐶 )

(10𝑔) (0.44 𝐽𝑔°𝐶 )+ (50𝑔 )(4 .18 𝐽

𝑔°𝐶 )

𝑇 𝑓=21.65 °𝐶