Presentation on theme: "Lets start with a new symbol Q. Q is for “Heat” Measured in Joules. Heat is a form of energy."— Presentation transcript:
Lets start with a new symbol Q. Q is for “Heat” Measured in Joules. Heat is a form of energy.
James Prescott Joule Discovered that heat was form of energy. Found that when water is stirred, temperature increases.
James Joule Experimented with falling masses turning paddles in water. Figured out how much mechanical energy is needed to change temperature. Got a unit named after him.
Units of Heat Joule is most common heat unit. Also, there is the calorie. Calorie: is the amount of energy necessary to raise the temperature of 1 g of water by 1° C. Units to measure heat. calorie: Energy needed to raise 1 g of water by 1C. 1 cal = Joules.
Calories and calories. Calories and calories are not the same thing. If C is uppercase, it is a “dietary calorie” (on your food label.) One Calorie = 1000 calories One Calorie = 1 kilocalorie One Calorie = 4186 Joules
Units of Heat, cont. Also – BTU BTU stands for British Thermal Unit A BTU is the amount of energy necessary to raise the temperature of 1 lb of water 1° F
New Symbol Specific heat = c Amount of heat needed to raise 1 kg of a material 1 C. Units J / kg °C Depends on the substance. Look it up. Specific heat questions are on the JCCC final.
Specific Heat equation Q = mcΔT Heat = mass*specific heat*Temp. Change If ΔT is positive, material gains heat. If ΔT is negative, material gives up heat.
Problem The Specific Heat of water is 4186 J/kgC. How much heat is needed to raise 6.2 kg of water from 10 to 85 Celsius? Do in PP notes. Specific heat questions like this are on the JCCC final.
Problem The Specific Heat of water is 4186 J/KgC. How much heat is needed to raise 6.2 kg of water from 10 to 85 Celsius? Q = mcΔT Q = 6.2kg*4186 J/KgC*75C
Calorimetry Measurement of heat flow from a hot material to a cold material (usually Water). Heat flow into water is calculated based upon the temp. change of the water. Conservation of energy applies to the isolated system. Don’t let any heat escape so that all goes into the water. (Insulated system) The energy that leaves the warmer substance equals the energy that enters the water. ΔQ hot = - ΔQ cold Know Q = mcΔT for final.
Easy problem Know for JCCC Final: ΔQ hot = - ΔQ cold mcΔT hot = mcΔT cold
Easy problem 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp?
Easy problem 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp? mcΔT hot = mcΔT cold 1 kg x 4186 J/kgK x (70C – T f ) = 1 kg 4186 J/kgK x (T f – 60C)
Easy problem 1 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp? 2 kg of water at 70 C is mixed with 1 kg of water at 60 C. What is the final temp?
Problem For breakfast you consume 320 dietary Calories. You decide to work this off by lifting a 25 kg barbell.4 m. How many times will you have to repeat this motion to burn off breakfast?
Problem For breakfast you consume 320 dietary Calories. Instead, you (60 kg) decide to spring from rest to 5 m/s. How many repetitions of this will you need to do?
Still talking about breakfast If instead of eating that food, you set it on fire, how much water could it bring from 0C to 100 C? Q = mcΔT m = Q /cΔT
What else does heat do? If heat is added to a substance, what else could happen besides ΔT?
Phase Change Temperature remains constant during phase change. E.g. Ice will continue to absorb heat at 0 C without changing temp. The temp will not be 1 C until all of the ice has melted. See graph next slide.
New Symbol l – “l” is for latent heat. Amount of heat absorbed, per kilogram, for a change from one phase to another. Heat absorbed during a phase change does not change temperature, so it is “latent.” Units: J/kg
Latent Heat, cont. Depends on both substance and phase change. Just something to look up. Latent heat of fusion, L f, is used for melting or freezing Latent heat of vaporization, L v, is used for boiling or condensing Latent heat of sublimation, L s, is used for sublimation or deposition.
Equation Q = m l
Problem A gram of H 2 0 is at -30 C. How much heat total is needed to raise it to H 2 0 at 120 C?
Multistep processes. A lot of problems will involve both a temperature change and a phase change. Evaluate the Q for each step separately, and then add them together.
Problem How many steps?
Graph of Ice to Steam How many steps? 5 1- Ice to 0 C 2- Melting 3- Water to 100 C 4- Boiling 5- Steam to 120 C Q total = Q 1 + Q 2 + Q 3 + Q 4 + Q 5
End here in 2012
Problem How many steps? 5 1- Ice to 0 C 2- Melting 3- Water to 100 C 4- Boiling 5- Steam to 120 C Q total = Q 1 + Q 2 + Q 3 + Q 4 + Q 5
Useful Information For H 2 0: c ice = 2090 J/kgº C l f = 3.33 X 10 5 J/kg c water = 4186 J/kgº C l v = 2.26 x 10 6 J/kg c steam = 2010 J/kgº C
Warm the Ice Q total = Q 1 + Q 2 + Q 3 + Q 4 + Q 5 Q 1 -> Ice from -30 to 0 Temp. change, so use specific heat. Q 1 =m ice *c ice *ΔT ice =.001kg* 2090 J/kgº C * ( )
Melt the Ice Q total = 63 J + Q 2 + Q 3 + Q 4 + Q 5 Q 2 -> Ice from 0 to 0 Phase change, so use latent heat. Q 2 =m ice *l fice =.001kg* 3.33 X 10 5 J/kg
Warm the Water Q total = 63 J J + Q 3 + Q 4 + Q 5 Q 3 -> Water from 0 to 100 Temp. change, so use specific heat. Q 3 =m water *c water *ΔT water =.001kg* 4186 J/kgº C * ( )
Boil the water Q total = 63 J J J + Q 4 + Q 5 Q 4 -> Ice from 0 to 0 Phase change, so use latent heat. Q 4 =m water *l fwater =.001kg* 2.26 x 10 6 J/kg
Warm the Steam Q total = 63 J J J J + Q 5 Q 5 -> Steam from 100 to 120 Temp. change, so use specific heat. Q 5 =m steam *c steam *ΔT steam =.001kg* 2010 J/kgº C * ( )
Warm the Steam Q total = 63 J J J J + 40 J Qtotal = 3115 J
Graph of Ice to Steam
Problem The latent heat of lead is J/kg, and the melting point is C. How much heat is needed to bring 17 kg of solid lead at C to 17 kg of liquid lead at C?
Problem Back in October, we learned that the most powerful person in the class was about 1050 Watts. If we gave him an electrical generator bike, how long would it take him to boil a kg of water from room temp?
Problem 400 g of Copper (c = 386 J/kgC) at 95 C is placed into a kilogram of water at 5 C. What will the final temperature of the mixture be?
Another problem. 10 kilograms of Aluminum (c = 900 J/kgC) are at 130 C. A 50 g Ice cube at -5 C is dropped onto the aluminum and melts, then boils away. What is the final temperature of the Al?
More than 2 things ΣΔQ = 0 Or ΔQ 1 + ΔQ 2 + ΔQ 3 … = 0
Problem 400 ml of water at 40 C is poured into a.3 kg glass beaker (c glass = 837 J/C) and a.5 kg block of Al at 37 C (c Al 900 J/C) is dropped in. What will the equilibrium temp be?