Including Temperature, Energy, Specific Heat Capacity, and Calorimetry Energy of Matter Including Temperature, Energy, Specific Heat Capacity, and Calorimetry

Temperature Temperature is a measure of the average kinetic energy of the particles in a sample of matter. (hmmm…why kinetic?) The hotter a substance feels or its temperature is, the more kinetic energy its particles have. The Kelvin and Celsius temperature scales are used in science.

Temperature Conversions The equation to convert Celsius to Fahrenheit: F = 1.8 °C + 32 What is the equation to convert Fahrenheit to Celsius? Write it down…

Temperature Conversions °F = 1.8 °C + 32 and °C = °F - 32 1.8 Convert the following: 45 °F into °C -50 °C into °F

Temperature Conversions The equation to convert Celsius to Kelvin: K = 273 + °C What is the equation to convert Kelvin to Celsius?

Practice temperature conversions Convert 25.00°C to Kelvin: Convert 300.00 K to Celsius: Convert -265.50°C to Kelvin:

Practice temperature conversions (Answers) Convert 25.00°C to Kelvin: K = 273 + °C K = 273 + 25.00°C = 298 K Convert 300.00 K to Celsius: °C = K - 273 °C = 300.00 K - 273 = 27 °C Convert -265.50°C to Kelvin: K = 273 + °C K = 273 + -265.50°C = 7.5 K

Energy Energy always flows from substances with higher temperature to substances with a lower temperature. Energy can be measured in calories or joules 1 calorie = 4.184 joules The caloric content of foods is actually kilocalories…so 1 Cal from food = 1 kcal.

Energy Transfer (try a simulation on pHet!)

Energy Conversions Convert 560.calories into joules: Convert 1400. joules into calories: Convert 250 Calories into calories:

Energy Conversions (answers) Convert 560.calories into joules: 560.0 cal x 4.184 J = 2343 cal 1 1 cal 2. Convert 1400. joules into calories: 1400. J x 1 cal = 334.6 cal 1 4.184 J Convert 250 Calories into calories: 250. Cal x 1000 cal = 2.50 x 10 5 cal 1 1 Cal

Specific Heat Capacity can be thought of as a measure of how much heat energy is needed to warm the substance up. You will possibly have noticed that it is easier to warm up a saucepan full of oil than it is to warm up one full of water. http://www.cookware-manufacturer.com/photo/418fa6490f24202f2cc5b5feee0fdde3/Aluminum-Saucepan.jpg

Specific Heat Capacity (C) or (SH) of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1oC (or by 1 K). The units of specific heat capacity are J/g oC or J/ g K . Because a difference of 1 degree on the Celsius scale is equivalent to a difference of 1 Kelvin on the Kelvin scale, either temperature scale can be used.

The next table shows how much energy it takes to heat up some different substances. The small values show that not a lot of energy is needed to produce a temperature change, whereas the large values indicate a lot more energy is needed. Which type of substances seem to have the lowest specific heat capacities???

Approximate values in J / g K or J/g°C of the Specific Heat Capacities of some substances: Air 1.000 Lead .129 Aluminum .897 Mercury .140 Asbestos .840 Nylon 1.700 Brass .400 Paraffin 2.100 Brick .750 Platinum .135 Concrete 3.300 Calcium .647 Cork 2.000 Polystyrene 1.300 Glass .600 Rubber 1.600 Gold .129 Silver .235 Ice 2.06 Steel .450 Iron .449 Water(l) 4.184 Copper .385 Ethanol(l) 2.44

The equation: q = m x SH x ΔT J g J/g °C °C The amount of heat energy (q) gained or lost by a substance = mass of substance (m) X specific heat capacity (SH) X change in temperature (ΔT) 
q = m x SH x ΔT J g J/g °C °C

What is ΔT? ΔT is called “delta T” Delta (Δ) means “change in” So ΔT is “change in temperature” When using ΔT in a problem you must identify the FINAL temperature and INITIAL temperature correctly. ΔT = TF - TI If ΔT is negative (-) the substance was cooling down

Example 1 of a calculation using the specific heat capacity equation: How much energy would be needed to heat 450. grams of copper metal from a temperature of 25.0ºC to a temperature of 75.0ºC? Note: (The specific heat of copper at 25.0ºC is 0.385 J/g ºC.)

Explanation: ΔT = Tfinal - Tinitial so 75. 0ºC - 25. 0 ºC = 50 Explanation: ΔT = Tfinal - Tinitial so 75.0ºC - 25.0 ºC = 50.0 ºC m = 450. g SH = 0.385 J/g ºC q = m x SH x ΔT . and plugging in your values you get q = (450. g) x (0.385 J/g ºC) x (50.0ºC) = 8700 J = 8.70 x 103 J (3 SF)

Example 2 (work this in class) What is the change in temperature of 515.00 g liquid water if 275,000 joules of heat are absorbed by it? BE: q = m x SH x ΔT RE: q= m= SH= ΔT=

Example 3 (work this in class) A piece of pure silver metal was cooled from 25.00°C to – 6.25 °C. To do this cooling, 175 kJ was required. What is the mass of the silver? BE: q = m x SH x ΔT RE: q= m= SH= ΔT=

Calorimetry Calorimetry is a lab technique that allows us to determine energy transfer between substances. Usually one substance is water and the other is a sample of some metal. Remember, energy always flows from warmer objects to cooler objects.

Calorimetry -qwarm = qcool WHY?? – (mm SHm ΔTm) = (mw SHw ΔTw) If the warm object transfers all of its energy to the cool object, then -qwarm = qcool WHY?? In our class the warm object is usually metal and the cool object is water, so we can use: – (mm SHm ΔTm) = (mw SHw ΔTw) You will have to rearrange the equation to solve for the variable you need.

Example 1 Calorimetry A 40.00 g piece of pure iron at a temperature of 78.0°C was dropped into a calorimeter containing water at 22.0°C. After a few minutes the temperature inside the calorimeter rose to 25.6°C. What is the mass of the water used? BE: – (mm SHm ΔTm) = (mw SHw ΔTw) Givens: mm = 40.00g mw = ? SHm = 0.449 J/g°C SHw = 4.184 J/g°C ΔTm= (25.6 °C - 78.0°C ) = -52.4 °C ΔTw= (25.6 °C - 22.0°C ) = 3.6°C

Example 1 Calorimetry BE: – (mm SHm ΔTm) = (mw SHw ΔTw) Givens: mm = 40.00g mw = ? SHm = 0.449J/g°C SHw = 4.184 J/g°C ΔTm= (25.6 °C - 78.0°C ) = -52.4 °C ΔTw= (25.6 °C - 22.0°C ) = 3.6°C RE: mw = – (mm SHm ΔTm) ( SHw ΔTw) Plug-n-chug: – (40.00g x 0.449J/g°C x -52.4 °C ) (4.184 J/g°C x 3.6°C ) mw = 62.4 g >>>>> 62 g water (2 SF)

Helpful hints for calorimetry problems ALWAYS list your givens, keeping metal givens separate from water givens. Make sure you have plenty of room on your paper to list givens and rearrange the basic equation. ΔT will ALWAYS be negative if the substance is cooling down…and positive if it is warming up. Want to check your work? The q of your metal should be nearly identical (but opposite in value) to the q of your water. Plug in values and try it! Tfinal will be the same for both the metal and the water, ALWAYS!! They are in the calorimeter together!

Example 2 Calorimetry (in class with your teacher) Callie heated a 83.00 g bar of aluminum metal and dropped it into a calorimeter containing 100.00 g water at 23.00 °C . The temperature inside the calorimeter rose to 32.00 °C . At what initial temperature must the aluminum have been? METAL – m WATER – w mm = cm = ΔTm = (Tf - Ti) mw = cw = ΔTw = (Tf - Ti)

– (mm cm ΔTm) = (mw cw ΔTw) Example 2 Calorimetry Callie heated a 83.00 g bar of aluminum metal and dropped it into a calorimeter containing 100.00 g water at 23.00 °C . The temperature inside the calorimeter rose to 32.00 °C . At what initial temperature must the aluminum have been? METAL – m WATER – w mm = 83.00 g cm = 0.897 J/g°C ΔTm = (Tf - Ti) = (32.00 °C – Ti) mw = 100.00 g cw = 4.184 J/g°C ΔTw = (Tf - Ti) = (32.00 °C – 23.00 °C) = 9.00 °C Basic Equation – (mm cm ΔTm) = (mw cw ΔTw) Rearranged Equation ΔTm = – (mw cw ΔTw) ( cm mm)

Rearranged Equation ΔTm = – (mw cw ΔTw) ( cm mm) ΔTm = – (100.00 g  4.184 J/g°C  9.00°C ) 83.00 g  0.897 J/g°C ΔTm = – 50.6 °C ΔTm = (Tf - Ti) – 50.6 °C = 32.00 °C – Ti + Ti & + Ti (add Ti to both sides) – 50.6 °C + Ti = 32.00 °C + 50.6 °C & + 50.6 °C (add + 50.6 °C to both sides) Ti = 82.6 °C (82.6 °C is the T of the metal before it was put in the calorimeter)