Chapter 12

Temperature and Heat

Younes Sina

Temperature

The temperature of an object is a measure of how cold or hot that

object is. More precisely, the temperature of an object is a measure of

the average kinetic energy of the atoms and molecules of that object.

A hotter object has faster molecular vibrations. Temperature is a scalar

quantity.

Temperature Scales

Fahrenheit and Rankin (in the American system of units)

Celsius and Kelvin (in the Metric System)

Fahrenheit and Celsius are regular scales.

Rankin and Kelvin are absolute scales.

Example : The Fahrenheit scale reads 77o F, what is the reading on

the Celsius scale?

Solution:

°C = (5/9)[ F - 32 ]oC = (5/9) [ 77 - 32] = 25

Example : A temperature difference of ΔC = 24oC is measured

between two points on Celsius scale. How much is this difference in

Fahrenheit scale?

Solution: 100oC difference on the Celsius scale corresponds to 180oF

difference on the Fahrenheit scale. Using a proportion, the difference

in Fahrenheit is

ΔF / ΔC = 180o / 100o

ΔF / 24o = 9/5

ΔF = 24o (1.8)

ΔF = 43oF.

C1 = (5/9)[ F1 - 32 ]C1 = (5/9)[ F1 - 32

C 2 =(5/9)[ F2 - 32 ]

(C1-C2)= (5/9)[(F1-32)-(F2-32)]ΔC=(5/9)ΔF

Example : At what temperature both Fahrenheit and Celsius scales read

the same temperature?

Solution:°C = (5/9)[ F - 32 ]

F = (5/9)[ F - 32 ]

F=(5/9)F-17.777

(4/9)F=-17.777F=-40

Absolute Scales:

The basis for absolute scales (Kelvin and Rankin scales) is the

temperature at which molecular motion ceases and stops. This

temperature cannot actually be reached; however, with great cooling,

temperatures very close to it have been reached. Experiments have

shown that when a gas is cooled down, its volume decreases.

At constant pressure, the volume decrease for a gas, is proportional to

the temperature decrease. In other words, The ratio ΔV/ΔT remains

constant. That means that if V( the gas volume) is plotted versus T

(the gas temperature) while pressure is kept constant, the graph is a

straight line as shown below:

HeatHeat is a form of energy that transfers due to a temperature difference.Units of HeatThe familiar unit often heard is "calorie."One calorie (1 cal) is the amount of heat energy that can raise the temperature of 1 gram of pure water by 1 oC.Parallel to this definition is that of kilocalorie (kcal).1 kcal is the amount of heat energy that can raise the temperature of 1 kg of pure water by 1 oC.A non-Metric unit for heat energy is Btu (British thermal unit).1 Btu is the amount of heat energy that can raise the temperature of 1 lbm of pure water by 1 oF.

Specific Heat (c)

Different substances take different amounts of heat energy for one unit

of mass of them to warm up by one degree. For example, if you place

1 kg of copper and 1 kg of aluminum on a burner, after the same length

of time, the copper piece will be much hotter than the aluminum. The

reason is the difference in their specific heat. Copper takes much less

heat to warm up compared to aluminum.

The specific heat (c) of a substance is the amount of heat that 1kg of

substance takes to warm up by 1oC. On this basis, the specific heat of

water is 1 kcal /(kg oC). The specific heat of a few elements are given

below:

Cwater =1.000 kcal / ( kg oC) or, cwater = 1.000 cal /(g oC)

ciron = 0.108 kcal / ( kg oC) or, ciron = 0.108 cal /(g oC)

cAl = 0.215 kcal / ( kg oC) or, cAl = 0.215 cal /(g oC)

Cbrass = 0.0924 kcal / ( kg oC) or, cbrass= 0.0924 cal /(g oC)

Heat calculation:

When heat is given to a pure substance or taken from it, its

temperature starts changing if a phase change does not start. During

a phase change (solid to liquid, liquid to solid, liquid to vapor, or

vapor to liquid ),temperature remains constant. We will look at the

following two cases:

1) Heat calculation for temperature change

2) Heat calculation for phase change

Heat Calculation When Temperature changes

Q = McΔT

No Phase Change

Mass of the objectspecific heat

temperature change

Example :

Calculate the amount heat that must be given to 2.14 kg of iron to

warm up from 24.0 oC to 88.0 oC.

Solution:Q = McΔT Q = ( 2.14kg )[0.108 kcal /(kg oC)] ( 88 - 24 )oC = 14.8 kcal

ΔE

Example :

37.0 cal of heat is given to 2.00 gram of water at 12.0 oC. Find its

final temperature.

Solution:

Q = McΔT

ΔT = Q / Mc

ΔT = (37cal) / [(2gr)(1cal/ gr oC)] = 18.5oC

Tf - Ti = 18.5 °C

Tf - 12.0 oC = 18.5 oC

Tf = 30.5 oC

Example :

Calculate the amount heat that must be taken from 5.00 kg of

Aluminum to cool it down from 230oC to 30oC.

Solution:

Q = McΔT

Q =(5.00kg )[0.215 kcal / ( kg oC )](30 - 230 )oC = - 215 kcal

Thermal Equilibrium

- [ heat loss by hotter objects ] = Heat gain by colder objects

Example :

A 65-gram piece of aluminum at 180 oC is removed from a stove and

placed in 45 grams of water initially at 22 oC. Find the equilibrium

temperature (Teq).

Solution:

-[ heat loss by hotter objects ] = heat gain by colder objects

Q = McΔT

-Mc [Tf - Ti] Al = Mc[Tf - Ti] water

- (65)(0.215)(Teq - 180) = (45)(1.00)(Teq - 22)

-14(Teq - 180) = (45)(Teq - 22)

-14Teq + 2520 = 45Teq - 990

3510 = 59Teq

Teq = 59 oC

Example :

A 225-gram piece of hot iron is removed from an electric oven at an

unknown temperature. It is known that the iron piece has been in the

oven long enough so that its initial temperature can be thought as the

temperature of the oven. The iron piece is placed in 75.0 grams of water

that is held by a 45.0-gram aluminum container initially at an equilibrium

temperature of 25.0 oC. The final equilibrium temperature of iron,

aluminum, and water becomes 41.0 oC. Find the initial temperature

of iron (oven). Assume that the whole system is thermally isolated

from the surroundings.

hot iron

225 gram

75.0 gram

water

45.0 gram

aluminum

25.0 oC

41.0 oC

Solution:-[ heat loss by hotter objects ] = heat gain by colder objects- Mc [Tf - Ti] iron = Mc [Tf - Ti] Al + Mc [Tf - Ti] water

-(225)(0.108)(41-Ti ) = (45)(0.215)(41-25) + (75)(1)(41-25)-24.3(41-Ti ) = 1354.8-41-Ti = 55.8Ti = 96.8 oC

Heat Calculation

Phase Change Only

During a phase change such as following, temperature remains constant: solid to liquidliquid to solidliquid to vaporvapor to liquid

Q = M Lf

latent heat of fusion

Lf is measured for different substances at their melting/freezing points or

temperatures. Typical values may be found in texts or handbooks.

Example : How much heat should be removed from 250 grams of wateralready at its freezing point (0 oC) to convert it to ice at (0 oC)? The latent heat of fusion (freezing) for water is Lf = 80 cal/gr.

Solution: In this problem, the heat calculation involves a phase change only. There is no temperature change.Q = M Lf

Q = (250gr)(80 cal/gr ) = 20,000 cal

0 oC 0 oC

Example :How much heat should be given to 250 gr of ice at 0oC to convert it to water at 40.oC ? The latent heat of freezing/melting for water is Lf = 80 cal /gr and the specific heat of water is 1 cal/[gr oC].Solution: In this problem, the heat calculation involves a phase change and a temperature change.Q = M Lf + Mc (Tf - Ti)Q = (250)(80) + (250)(1)(40 - 0) = 30,000cal

-20 oC0 oC

250 gr

100 oC

Example :How much heat should be given to 250 gr of ice at -20 oC to bring it to boil (100 oC)?

0 oC

Solution:This problem has 3 steps:A temperature change ( ice from -20oC to 0oC )A phase change ( ice at 0oC to water at 0°C )A temperature change ( water from 0oC to water at 100oC)cice = 0.48 cal / (gr oC)Lf = 80 cal/grcwater = 1cal / (gr oC)Q = Mcice (Tf - Ti) + M Lf + Mcwater (Tf - Ti)Q = (250)(0.48)[ 0 - (-20) ] + (250)( 80) + (250)( 100 - 0) = 47,000 cal

2400cal 20,000cal 25,000cal

phase changetemperature change temperature change

Example :Draw a diagram that shows temperature change vs. heat consumed for previous Example.

Solution:

Example :2650 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final temperature of the result.

Solution: All of the ice may not melt. Let's first calculate the heat necessary for the ice to warm up to 0 oC-ice.Q = Mc ( Tf - Ti ) = (225gr)(0.480 cal/gr oC)[ 0 - (-12.0)]oC = 1300 calThe remaining heat is 2650cal - 1300cal = 1350 calNow, let's see if this heat is enough to melt the iceQ = MLf = (225gr)(80. cal /gr ) = 18000 calNo, 1350 cal is not enough to melt all of the ice. We need to see how much of the ice does melt.Q = MLf

1350cal = M (80 cal / gr)M = 17 grams.The final result is a mixture of17 grams of water and (225 - 17) gramsof ice, of course both at 0oC.

Example :26500 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final temperature.

Solution:All of the ice may not melt. First calculate the needed heat for the ice to warm up to 0 oC-ice.Q = Mc ( Tf - Ti ) = (225gr)(0.480cal /(gr oC)[ 0.0 - (-12.0)]oC = 1300 calThe remaining heat is 26500cal - 1300cal = 25200 calNow, let's see if this heat is enough to melt the ice.Q = MLf = (225gr)(80cal /gr ) = 18000 calYes, 25200 cal is enough to melt all of the ice.The remaining heat is 25200cal - 18000cal = 7200 cal. This heat warms up water from 0oC to Tf .Q = Mc ( Tf - Ti )7200cal = (225gr)(1 cal/gr oC)( Tf - 0oC )Tf - 0.0oC = 7200 / 225 = 32oCTf = 32oCThe final result is 225 grams of water at 32oC.

Homework : problems

chapter 12