Chapter 6, Section 8 Energy Consumption: Cold Shower.

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Presentation transcript:

Chapter 6, Section 8 Energy Consumption: Cold Shower

HW: PTG 6.8 p 686 #1-8, Due Friday Learning Objective – Calculate the energy efficiency of electrical devices. Success Criteria – Calculate the heat gained by a sample of water. – Calculate the electrical energy converted into heat by a resistor. – Calculate the efficiency of a transformation of electrical energy to heat. – Explore the power ratings and energy consumption levels of a variety of electrical appliances. Learning Objective – Calculate the energy efficiency of electrical devices. Success Criteria – Calculate the heat gained by a sample of water. – Calculate the electrical energy converted into heat by a resistor. – Calculate the efficiency of a transformation of electrical energy to heat. – Explore the power ratings and energy consumption levels of a variety of electrical appliances. Do Now – 6.7 Quiz – LO and SC on new left side page – WDYS/WDYT p 678 Agenda – Do Now – Investigation – Summary Do Now – 6.7 Quiz – LO and SC on new left side page – WDYS/WDYT p 678 Agenda – Do Now – Investigation – Summary

What do you see?

What do you think? The entire daily energy output of a Homes For Everyone (HFE) generator would not be enough to heat water for an average American family for a day. If an electrical heating coil (a type of resistor) were submerged in a container of water, and if a current were to flow through the coil to make it hot, what factors would affect the temperature increase of the water? The entire daily energy output of a Homes For Everyone (HFE) generator would not be enough to heat water for an average American family for a day. If an electrical heating coil (a type of resistor) were submerged in a container of water, and if a current were to flow through the coil to make it hot, what factors would affect the temperature increase of the water?

Investigation 6.8 #1 – setup and measurements of water and coil #2 Heat water for 10 minutes While water heats set up notebook #1 – setup and measurements of water and coil #2 Heat water for 10 minutes While water heats set up notebook

Investigation 6.8 Recall that the heat gained by an object can be found using the equation: ΔQ = mcΔT – ΔQ is a measure of heat in joules, – m is the mass of the substance in grams – c is the specific heat of the substance, and – ΔT is the change in temperature in degrees Celsius. Recall that the heat gained by an object can be found using the equation: ΔQ = mcΔT – ΔQ is a measure of heat in joules, – m is the mass of the substance in grams – c is the specific heat of the substance, and – ΔT is the change in temperature in degrees Celsius. #3 – Calculate the heat gained using your data table and the specific heat value of4.18 J/g°C.

Investigation 6.8 P=E/t so E=Pt where E is energy in joules (watt-seconds), P is power in watts (joules/second), t is time in seconds. P=E/t so E=Pt where E is energy in joules (watt-seconds), P is power in watts (joules/second), t is time in seconds. #4 - Calculate the energy from your experiment using this equation Answer 4 a and 5b #4 - Calculate the energy from your experiment using this equation Answer 4 a and 5b

Investigation 6.8 #5 – Calculate the efficiency using the equation above and answer 5a

Summary

HW: PTG 6.8 p 686 #1-8 Learning Objective – Calculate the energy efficiency of electrical devices. Success Criteria – Calculate the heat gained by a sample of water. – Calculate the electrical energy converted into heat by a resistor. – Calculate the efficiency of a transformation of electrical energy to heat. – Explore the power ratings and energy consumption levels of a variety of electrical appliances. Learning Objective – Calculate the energy efficiency of electrical devices. Success Criteria – Calculate the heat gained by a sample of water. – Calculate the electrical energy converted into heat by a resistor. – Calculate the efficiency of a transformation of electrical energy to heat. – Explore the power ratings and energy consumption levels of a variety of electrical appliances. Do Now – List some ways to make the foam cup setup more energy efficient. Agenda – Do Now – Investigation – Summary Do Now – List some ways to make the foam cup setup more energy efficient. Agenda – Do Now – Investigation – Summary

Physics Talk 6.8 What is a kilowatt- hour? 1 kilowatt-hour of energy power time 1000W x 1 h = 1000W x 3600 s = 3,600,000 J That’s a lot of joules! 1 kilowatt-hour of energy power time 1000W x 1 h = 1000W x 3600 s = 3,600,000 J That’s a lot of joules!

Physics Talk 6.8 What is heat? Heat is the transfer of thermal energy. In order for an object to heat up, there must be a source of thermal energy When objects cool down, the thermal energy is being transferred elsewhere Heat is the transfer of thermal energy. In order for an object to heat up, there must be a source of thermal energy When objects cool down, the thermal energy is being transferred elsewhere

Physics Talk 6.8 How does a thermos work? A thermos creates a barrier to isolate the contents. This means that the amount of energy in the thermos must stay relatively the same. Cold things stay cold (blocks thermal energy from leaving) Hot things stay hot (traps thermal energy in) A thermos creates a barrier to isolate the contents. This means that the amount of energy in the thermos must stay relatively the same. Cold things stay cold (blocks thermal energy from leaving) Hot things stay hot (traps thermal energy in)

Physics Talk 6.8 How do we measure a change in heat? ΔQ = mcΔT – ΔQ is a measure of heat in joules, – m is the mass of the substance in grams – c is the specific heat of the substance, and – ΔT is the change in temperature in degrees Celsius. ΔQ = mcΔT – ΔQ is a measure of heat in joules, – m is the mass of the substance in grams – c is the specific heat of the substance, and – ΔT is the change in temperature in degrees Celsius.

Physics Talk 6.8 Another method for calculating a change in heat We know energy = power x time We also know power = voltage x current This means, energy = voltage x current x time E = VIt where E is the energy in joules (J) V is the voltage in volts (V) I is the current in amperes (A) t is the time in seconds (s) We know energy = power x time We also know power = voltage x current This means, energy = voltage x current x time E = VIt where E is the energy in joules (J) V is the voltage in volts (V) I is the current in amperes (A) t is the time in seconds (s)

Sample Problem A 12-V starter battery in a car supplies 48 A of current to the starter. If the starter draws energy for 15 s, how much energy does the starter use? Given: – V = 12V – I = 48A – t = 15s Unknown: – Energy! (E) Tool: – E = VIt Solution: – E= 12V·48A·15s – E = 8640 J Given: – V = 12V – I = 48A – t = 15s Unknown: – Energy! (E) Tool: – E = VIt Solution: – E= 12V·48A·15s – E = 8640 J

Canceling Units

Physics Talk 6.8 What happened to the missing energy in the experiment? Some of the heat could have been lost out of the cup, into the thermometer, or somewhere else. The important thing is that it went somewhere because ENERGY IS NEVER DESTROYED! Some of the heat could have been lost out of the cup, into the thermometer, or somewhere else. The important thing is that it went somewhere because ENERGY IS NEVER DESTROYED!

Physics Talk 6.8 What is the equation for efficiency?

Essential Questions P 684 WDYTN Questions p 685 – What does it mean? – How do you know? – Why do you believe? – Why should you care? P 684 WDYTN Questions p 685 – What does it mean? – How do you know? – Why do you believe? – Why should you care?

Physics To Go #5-8 as a class