Temperature Conversions

Temperature

• Definition:• Temperature is the average kinetic energy, KE,

of molecules in a sample. • Absolute zero is the temperature at which

molecules stop all motion.• (PE – potential energy is stored energy)• (KE – kinetic energy, energy of motion)

Temperature Conversions

C FF = 9/5C + 32F CC = 5/9(F-32)

C KK = C + 273K CC = K - 273

Comparing Temperature Scales

212

Temperature ConversionsC. Do the following problems on another sheet of paper. Use the conversion

equations from above. Show all your work: Knowns & Unknowns, Set-Up & Answer (K = C + 273, F = 9/5C + 32)

1. Convert the following centigrade temperatures into fahrenheit.a. 21 'C (A: 70'F) b. -10 'Cc. 300 'C d. -78.5 'C

2. Convert the following centigrade temperatures into kelvin.a. 21 'C (A: 294 K) b. -10 'Cc. 300 'C d. -78.5 'C

3. Convert the following fahrenheit temperatures into celcius.a. 75 'F (A: 24 'C) b. -10 'Fc. 100 'F d. 212 'F

4. Convert the following kelvin temperatures into celcius.a. 77 K (A: -196 'C) b. 198.5 Kc. 373 K d. 500 K

Part C Temperature Conversions Answer Key

A B C D

1 70F 14F 572F -109F

2 294K 263K 573K 195K

3 24C -23C 38C 100.C

4 -196C -75C 100.C 227C

Temperature Quiz(include problem, work & answer)

1. 456 K = ? CC = K – 273C = 456 – 273C = 183C

2. 456 C = ? FF = 9/5C + 32F = 9/5(456) + 32F = 853F (round to nearest 1 due to

adding 32)

Heat & Calorimetry

Heat (Q) – is defined as the kinetic energy associated with the motion of countless atoms and molecules which can be transferred between objects of different temperature.

Temperature (T) – the average kinetic energy of the atoms or molecules in a sample of matter.

Calorimetry – measurement of the amount of heat gained or lost in a chemical reaction.

Calorimetry CalculationsEquation: q = Cp x m x DT

q = heatCp = specific heat (heat capacity), the amount of

heat needed to raise up the temperature of 1 g of the specific substance 1 degree C.

m = mass of sample testedDT = change of temperature of sampleSolve the equation for Cp:

Cp = q / (m DT )

Specific Heat Lab

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

Do 3 trials for each substance: wood, copper, glass, silver, sodium & water

Make 2 tables for each substance: Observations & Data Table, Calculations Table

Calculate Cp for all 6 substances with 3 trials each for a total of 18 calculations.

Observations & Data

Material____________ Trial #1 Trial #2 Trial #3

Mass of object, m (g) g g g

Initial temperature of object (T1)

ºC ºC ºC

Final temperature of object (T2)

ºC ºC ºC

Energy absorbed (Joules) J J J

Calculations:

Material____________ Trial #1 Trial #2 Trial #3

ΔT = (T2 - T1) ºC ºC ºC

Cp = q / (mDT) J/(g · ºC) J/(g · ºC) J/(g · ºC)