Heat Capacity and Specific Heat Exercise Question

Lecture 24 Purdue University, Physics 220 1

Lecture 24

Heat

PHYSICS 220


Heat Capacity and Specific Heat Heat capacity =Q/!T

• shows how much heat is required to change the

T of object (system)

• Specific heat c = Q/m!T

• Q = c m !T

Heat required to increase temperature depends on

amount of material (m) and type of material

• Heat adds energy to object/system

• IF there is no dissipation then:

Heat increases internal energy: Q = !U

Heat increases temperature: Q = C !T


Exercise

After a grueling work out, you drink a liter of cold

water (0 C). How many Calories does it take for

your body to raise the water up to body

temperature of 36 C?

A) 36 B) 360 C) 3,600 D) 36,000

1 liter = 1,000 grams of H20

1000 g x 1 calorie/(gram degree) x (36 degree) = 36,000 calories

36,000 calories = 36 Calories!


Question

Suppose you have equal masses of aluminum and

copper at the same initial temperature. You add 1000 J

of heat to each of them. Which one ends up at the

higher final temperature

A) aluminum

B) copper

C) the same

!T = Q/cm

Substance c in J/(kg-C)aluminum 900copper 387iron 452lead 128human body 3500water 4186ice 2000


Specific Heat for Ideal Gas

• Monatomic Gas (single atom)

Translational kinetic energy only

At constant Volume work = 0

Q = !U = 3/2 nR!T

CV = 3/2 R = 12.5 J/(K mole)

Cv – specific heat at constant volume.

• Diatomic Gas (two atoms)

Can also rotate

CV = 5/2 R = 20.8 J/(K mole)


Phase Transitions

• A phase transition occurs whenever a material is changed

from one phase, such as the solid phase, to another

phase, such as the liquid phase.

– Phase transitions occur at constant temperature.

– The latent heat of vaporization LV is the heat per unit mass that

must flow to change the phase from liquid to gas or from gas to

liquid.

• Fusion occurs when a liquid turns into a solid.

• Evaporation occurs when a liquid turns into a gas.

• Sublimation occurs when a solid changes directly to a gas

without going into a liquid form.

Demo 3B - 04

T pinned at boiling

point of water

which is below

ignition point

for cup


• As you add heat to water, the temperature increases

for a while, then it remains constant, despite the

additional heat!

• Latent Heat L [J/kg] is heat which must be added (or

removed) for material to change phase (liquid-gas).

• |Q| = m L

Latent Heat

T

Q added to water

water temp rises

water changesto steam

(boils)

steam temp rises

100oC

Latent HeatSubstance Lf (J/kg) Lv (J/kg)

water 33.5 x 104 22.6 x 105

f=fusion v=vaporization

WhyWhen boiling the most energetic molecules escape

Reducing the average T

But heat added to continue boiling so

process goes to completion


Phase Diagram

H2O


Phase Diagram

CO2


Exercise

During a tough work out, your body sweats (and evaporates)

1 liter of water to keep cool (37 C). How much cold water

would you need to drink (at 2 C) to achieve the same thermal

cooling? (recall CV = 4.2 J/g for water, Lv=2.2x103 J/g)

A) 0.15 liters B) 1.0 liters C) 15 liters D) 150 liters

Qevaporative = L m = 2.2x103 kJ/kg x 1kg

Qc = c m !t = 4.2kJ/kgK x 35K x m

m = 2.2x103 / 147 = 15kg or 15 liters!


Boiling Point

Going from Lafayette to Denver the

temperature at which water boils:

A) Increases B) Decreases C) Same


Exercise

How much ice (at 0 C) do you need to add to 0.5 liters of awater at 25 C, to cool it down to 10 C?(L = 80 cal/g, c = 1 cal/g C)

Qwater

= mc!T

= (0.5kg)(1cal / gC)(15C)

= (7,500 calories)

Qice= mL + mc!T

" m =Q

ice

L + c!T

=7,500cal

80cal / g + (1cal / gC)(10)= 83.3 grams

Not same m


Exercise

Ice cube trays are filled with 0.5 kg of water at 20 C andplaced into the freezer. How much energy must beremoved from the water to turn it into ice cubes at -5 C?(L = 80 cal/g, cwater = 1 cal/g C, cice = 0.5 cal/g C)

Q1= mc

water!T

1

= 500 "1" (#20) = #10000(cal)

Water going from 20 C to 0 C:

Water turning into ice at 0 C:

Q2= !mL

= !500 " 80 = !40000(cal)

Ice going from 0 C to -5 C:

Q3= mc

ice!T

2

= 500 " 0.5" (#5) = #1250(cal)

! Q = Q

1+Q

2+Q

3= "51250(cal)

Heat transfer

• Conduction

• Convection

• Electro-magnetic radiation


Demo 3B - 03

Char were little conduction


Heat Transfer: Conduction

• Hot molecules have more KE than

cold molecules

• High-speed molecules on left

collide with low-speed molecules on

right

– energy transferred to lower-speed

molecules

– heat transfers from hot to cold

– vibrations


Heat Transfer: Conduction

• I = rate of heat transfer = Q/t [J/s]

I = " A (TH-TC)/d

• Q/t = " A !T/!x

" = thermal conductivity

• Units: J/s*m*C

• good conductors…high ", e.g., metal

• good insulators … low ", e.g., plastic

R = d/(A") = thermal resistance

TH

Hot

TC

Cold

d = !x

Area A


Conduction

Which of the following is an example of conductive heat

transfer?

A) You stir some hot soup with a silver spoon and notice that

the spoon warms up.

B) You stand watching a bonfire, but can’t get too close

because of the heat.

C) Its hard for central air-conditioning in an old house to cool

the attic.


• Find I=Q/t in J/sKey Point: Continuity (just like fluid flow)

• I1 = I2• "1A(T0-TC)/!x1 = "2A(TH-T0)/!x2

• solve for T0 = temp. at junction

• then solve for I1 or I2• TH-T0 = I R1 and T0-TC = I R2

!T = (TH-T0) + (T0-TC) = I (R1 + R2)

!x1 = 0.02 m A1 = 35 m2 k1 = 0.080 J/s*m*C

!x2 = 0.075 m A2 = 35 m2 k2 = 0.030 J/s*m*C

answer: T0=2.27 C I=318 Watts

Inside: TH = 25COutside: TC = 0C

I1 I2

T0

Conduction with 2 Layers

Documents

Heat Capacity and Specific Heat Exercise Question