Heat and temperature

April

Vocabfor review

Celsius condensation conduction convection energy transfer evaporation freezing gas heat joules Kelvin kinetic energy law of conservation of energy liquid melting molecular level phase phase change potential energy pressure radiation solid specific heat sublimation temperature thermal energy work

Vocab for mastery

law of entropy thermal equilibrium thermodynamics

Non-academic vocab

closed system co-efficient component contraction decay deterioration disorder efficient expansion macroscopic medium / media relate directly

Engagement

http://jersey.uoregon.edu/vlab/Thermodynamics/index.html

*Thermal equilibrium: the state in which two bodies in physical contact with each other have identical temperatures.

*

Engagement activity

EXPLORATION

• Virtual Lab – Determination of Specific Heat• The amount of heat required to raise the

temperature of a solid body depends on its change in temperature (DeltaT), its mass (m), and an intrinsic characteristic of the material forming the body called specific heat (cp). The heat is calculated from the equation

• q = cp x m x ΔT

• Historically, heat (q) was measured in terms of calories. The calorie was defined as the amount of heat required to raise the temperature of 1 gram of water by 1 ⁰C from 14.5 ⁰C to 15.5 ⁰C at 1 atmosphere pressure. With this definition, the specific heat of water is 1.00 cal/(g · ⁰C ). The use of the calorie began before it was established that heat is a form of energy and 1 calorie is equivalent to 4.18 J. The joule (J) has become the more favored unit in recent years. Thus, the units for cp that we will use are J/(g · ⁰C ). The specific heat of water is then 4.18 J/(g · ⁰C ).

• PURPOSE:• To apply the experimental methods of

calorimetry* in the determination of the specific heat of a metal.

• *calorimetry: an experimental procedure used to measure the energy transferred from one substance to another as heat.

• PROCEDURE:• Perform three trials for EACH material. Record the mass of

the sample, the initial temperature (T1), the final temperature (T2), and the energy absorbed (in joules).

• Vary the mass of the sample and the length of time heating for each trial with a given material. In other words, no two trials for a given material should use both the same mass and the same flame exposure times.

• At the end of each trial, hit the "Reset" button to prepare for the next trial.

• http://www.sciencegeek.net/VirtualLabs/SpecificHeatLab.html

RESULT

Material___________

Trial #1 Trial #2 Trial #3

Mass of object, m (g)

g g g

Initial temperature of object (T1)

⁰C ⁰C ⁰C

Initial temperature of object (T2)

⁰C ⁰C ⁰C

Energy absorbed (Joules)

J J J

• Observation and data:

result

• CALCULATIONS: Show your work! You will repeat the calculations three times for each material you test.

• Calculate the temperature change, ΔT, (T2 - T1).

• Calculate the specific heat, Cp. Cp = q ÷ m ÷ ΔT

Material______________

Trial #1 Trial #2 Trial #3

ΔT, (T1 – T2) ⁰C ⁰C ⁰C

Cp j/(g · ⁰C) j/(g · ⁰C) j/(g · ⁰C)

Temperature scales

• Celsius and Fahrenheit scales• Tf=9/5 Tc + 32

• Celsius and Kelvin (or absolute) scales• T= Tc + 273.15

practice

• One day it was -40 ⁰C at the top of Mont Blanc and -40 ⁰F at the top of Mount Whitney. Which place was colder?

• Neither (-40 ⁰C =-40 ⁰F)

temperature and thermal equilibrium

• When it is 75⁰F, you feel warm. When it is 45 ⁰F, you feel cool.

• Air has more energy at 75 ⁰F than it has at 45 ⁰F. This energy is the kinetic energy of air molecules.

• When two liquids or gases at different temperatures are mixed, the stable temperature is between the two temperatures. This is called thermal equilibrium.

conceptual challenge

• If two cups of hot chocolate,, one at 50 ⁰C and the other at 60 ⁰ C, are poured together in a large container, will the final temperature of the double batch be

• a. less than 50 ⁰ C?• b. between 50 ⁰F and 60 ⁰ C?• c. greater than 60 ⁰ C ?

heat and energy

• Why do you feel colder with the tile floor than with the wooden floor?

• The tile transfers or conducts energy faster than the wood does. More energy is removed from your feet in a given time.

• That’s why you feel colder with the tile floor.

• Heat always moves from an object at higher temperature to an object at lower temperature.

• Heat is defined as energy in transit.

• Heat has the units of energy.

• Joule (J)• calorie (cal)=4.186J

Specific heat capacity

• We experience that an object may be hotter than the other in the same condition.

• That property of objects is known as specific heat capacity.

• The specific heat capacity is defined as the energy required to change the temperature of 1 g of a substance by 1 ⁰C.

specific heat capacity

• cp = q / mΔT• Specific heat capacity = heat energy/(mass x

change in temperature)• Water: 4.186 (J/g ⁰C) • Iron: 4.48 (J/g ⁰C)

calorimetry

• You are preparing to take a bath. The cold-water faucet supplies water at 20.0 ⁰C, and the water from the hot-water faucet is 60.0 ⁰C. Each faucet has poured 25.0 kg of water into the tub. What is the temperature of the bath?

• 40.0 ⁰C

Calorimetry practice

• You prefer your bath at 30.0 ⁰C. The hot-water faucet has already poured 20.0 kg of water at 60.0 ⁰C into the tub. How much cold water (20.0 ⁰C) should you add?

• 60.0kg

Practice

• What is the final temperature when a 3.0 kg gold bar at 99 ⁰C is dropped into 0.22 kg of water at 25 ⁰C?

• 1. set the final temperature as T.• 2. the gold bar loses heat and water gains heat. • 3. loss of heat is q =cp mΔT = 0.129x3000x(99-T)

• 4. gain of heat is q= cp mΔT =4.186x220x(T-25)• 5. Loss of heat is equal to gain of heat. • 0.129x3000x(99-T) = 4.186x220x(T-25)• 387(99-T)=920.92(T-25)• T=47 ⁰C

Homework

• Page 374• Practice 10C• Questions 2, 3 and 7• For the specific heat

capacities of substances, refer to the table 10-4 in page 372. Be aware that the specific heat in the textbook is measured in J/kg ⁰C.

• Answers• 18.0 ⁰C• 79 ⁰C• 135 g

Phase change

1. What is the state of substances during Leg “A”?

2. What is the state of substances during Leg “B”?

3. What is the state of substances during Leg “D”?

• 1. Leg “A”: temperature of • ice changes as energy is • added.• 2. Leg “B”: temperature of

ice does not change. Instead, the nature of the ice changes. The ice-and-water mixture remains at this temperature until all of the ice melts.

3. Leg “C”: the water’s temperature increases as energy is added.

• 4. Leg “D”: temperature• stops rising and the water• turns into steam.• 5. Leg “E”: the steam’s

temperature increases.• During Leg “B” and Leg “D,” the energy added

or removed changes the internal energy of the substance without changing its temp. These changes in matter are called phase changes.

Phase change

vaporization

condensation

meltingsolidification

Phase change

heat of vaporization 2.26x103 J/g

heat of fusion 3.33 x102 J/g

heat of phase change practice• How much energy is required to change a 42 g ice cube from ice at

-11 ⁰C to steam at 111 ⁰C?• (heat of vaporization is 2.26x103 J/g, and heat of fusion is 3.33 x102

J/g)• 1. at the phase of ice: q=cp mΔT=4.186x42x11• 2. at the state of a mixture of ice and water: heat of fusion x mass=

3.33 x102 J/g x 42g• 3. at the phase of water: q=cp mΔT =4.186x42x100• 4. at the state of a mixture of water and steam: heat of

vaporization x mass = 2.26x103 J/g x 42g• 5. at the phase of steam: q=cp mΔT =4.186x42x11• Answer: 1.30 x 105 J

Draw a phase change diagram

• Ice at -20⁰C is heated to steam at 133.0 ⁰C.

Homework practice

• Page 381 • #3, #4 and #5

• Answers• 1.415 x104J• 1.354 x107J• 76.2 ⁰C