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Kinetic Energy Energy an object possesses when in motion. Law of Conservation of Energy – in any physical or chemical change, energy can change form,

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Presentation on theme: "Kinetic Energy Energy an object possesses when in motion. Law of Conservation of Energy – in any physical or chemical change, energy can change form,"— Presentation transcript:

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2 Kinetic Energy Energy an object possesses when in motion. Law of Conservation of Energy – in any physical or chemical change, energy can change form, but the total amount of energy remains constant.

3 HEAT Energy in a system can be transferred as heat. Heat – the sum total of kinetic energy of the particles in a sample of matter.

4 Temperature A measure of the average kinetic energy of particles in a sample of matter. Temperature is used to measure changes in heat energy

5 Temperature Scales Temperature is commonly measured using the Kelvin (K) and the Celsius ( o C) scales. To convert from one scale to another, the following equations can be used: o C = K – 273 K = o C + 273

6 Specific Heat Not all substances will experience the same changes in temp with equal additions of heat energy. This is due to a substance’s specific heat capacity. (Cp) Cp – the amount of energy required to raise the temp of 1 gram of a substance by 1 degree Kelvin.

7 Specific Heat (Cp) Cp = heat lost or gained (J or cal) mass (g) x change in temp (K or o C) ***Note: The units for heat are Joules (J) or the calorie (cal) A substance’s specific heat can be calculated using the following equation:

8 Specific Heat Problem In a lab experiment, 58.1 g of copper loses 1600.o J in cooling from 99.8°C to 28.4°C. Calculate the specific heat. Cp = _____(1600.0 J)______ (58.1 g) x (99.8°C – 28.4°C) Cp = _____(1600.0 J)______ (58.1 g) x (71.4°C) Cp = 0.386 J / g-°C Cp = heat lost or gained (J or cal) mass (g) x change in temp (K or °C)

9 Specific Heat Practice Problem How much heat energy is required to raise the temperature of 44.0 g of lead from 21°C to its melting point, 327°C? The specific heat of lead is 0.13 J/g-°C. Solve for heat energy 0.13 J/g-°C = _____[heat]______ (44.0g) x (327°C - 21°C) 0.13 J/g-°C = _____[heat]______ (44.0g) x (306°C) Heat = (0.13 J/g-°C) x (44.0g) x (306°C) Heat = 1750.32 J (**the g and °C cancel out!) Cp = heat lost or gained (J or cal) mass (g) x change in temp (K or °C)

10 Temperature Practice! 1)Write down the two equations for converting between Kelvin and Celsius. Convert the following: Write problems and answers for full credit! 2) 300 K = ______ºC 8) 0ºC = _______ K 3) 215 K = ______ºC9) 173 K = ______ºC 4) 81 ºC = _____K 10) 84ºC = _______ K 5) 32 ºC = _____K 11) -20ºC = ______K 6) 0 K = ______ ºC12) -100ºC = _____ K 7) 100 ºC = _____ K


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