Chapter 10: Energy. What is happening when an ice cube melts in your hand? How do fireworks work? What is energy? (answer on space provided on handout)

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Presentation transcript:

Chapter 10: Energy

What is happening when an ice cube melts in your hand? How do fireworks work? What is energy? (answer on space provided on handout)

Heat Energy flow caused by a temperature difference Measured in J or cal 1 cal = J Temperature Measure of random motion of a substance Measured in ̊ C, ̊ F, or K

System vs. Surroundings A burning match Energy is flowing _________ the system therefore the reaction is _____thermic. Melting ice cube Energy is flowing _________ the system therefore the reaction is _____thermic.

Calculations Convert temperature Change in temperature Convert to grams of a substance Q=SM∆T

Converting T ( ̊ F to ̊ C) Fahrenheit to Celsius T ( ̊ C) = ( T(°F) – 32) x (5/9) Ex: Convert 65 ̊ F to ̊ C T ( ̊ C) = ( T(°F) – 32) x (5/9) T ( ̊ C) = ( 65 °F – 32) x (5/9) T ( ̊ C) = 18. ̊ C

Convert T (K to ̊ C) Kelvin to Celsius T ( ̊ C) = T(K) – 273 Ex: Convert 300. K to C T ( ̊ C) = 300. K – 273 T ( ̊ C) = 27. ̊ C

Change in T (∆T) ∆T = T final – T initial Ex: If a cup of boiling water is left on the counter until it comes to room temperature (20 ̊ C), what is the change in temperature? ∆T = T final – T initial ∆T = 20 ̊ C – 100 ̊ C ∆T = -80 ̊ C

Convert to grams (from mol) Convert 6.00 mol of H 2 O to grams mol H 2 O18.02 g H 2 O = 1 mol H 2 O 108. g H 2 0

Q = SM∆T Q = Energy (heat) required in Joules S = specific heat capacity in J/g ̊ C M = mass of the sample in grams ∆T = change in temperature in ̊ C

SubstanceSpecific Heat Capacity (J/g ̊ C) air1.00 carbon (graphite)0.711 carbon dioxide0.832 granite0.803 iron0.448 lead0.128 mercury0.14 water (liquid)4.184 ice2.03 steam1.99 Specific Heat Capacity The amount of energy required to raise the temperature of one gram of a substance by one degree Celsius.

Q = SM∆T How much energy is needed to warm 7.40 g of water from 17.0 ̊ C to 35.0 ̊ C? Q = want S = J/g ̊ C (from table) M = 74.0 g ∆T = 35.0 ̊ C – 17.0 ̊ C= 18.0 ̊ C Q = SM∆T Q = (4.184 J/gC)(7.40 g)(18.0 C) Q = 557 J

Q = SM∆T If 72.4 j of heat is applied to a 9.52 g black of metal, the temperature increases by 52.3 F. Calculate the specific heat capacity. Q = S = M = ∆T = Q = SM∆T

Calculate the amount of energy required (in calories) to heat 145 g of water from 22.3 ̊ C to 75.0 ̊ C. Q = S = M = ∆T = Q = SM∆T

Lab Title, Data, Analysis, and Conclusion need to be in lab notebook

Reading 10.2 (pg 291 – 292) 10.5 (pg 294 – 300) Practice Problems Pg # 6-10, 22-28, 49-58