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Specific Heat

Specific Heat. Chemistry Music Video 4: Em Cee Delta Tee - YouTube. Specific Heat…. …is defined as the amount of heat needed to raise the temperature of 1 gram of a substance 1 K or 1 º C ….is measured in J/g ·K

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Specific Heat

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  1. Specific Heat Chemistry Music Video 4: EmCee Delta Tee - YouTube

  2. Specific Heat… • …is defined as the amount of heat needed to raise the temperature of 1 gram of a substance 1 K or 1º C • ….is measured in J/g·K • … will be lower for metals which means it takes very little heat to raise its temperature. • …of water is relatively high at 4.18 J/g· K which means …..

  3. Specific Heat • Basic equation: Q= mcpΔT, where Qrepresents heat energy, measured in Joules or kJ m represents mass in grams or kg cp represents specific heat in J/g· K Can also be kJ/kg· K ΔT represents temperature change, in K or ºC ΔT means (Tf - Ti) Rearrange the above equation for cp.

  4. 1. A 4.0 gram sample of glass was heated from 274.0 K to 314.0 K and was found to absorb 32 J of heat energy. Calculate the specific heat of this glass. • Givens: BE: RE:

  5. A 4.0 gram sample of glass was heated from 274.0 K to 314.0 K and was found to absorb 32 J of heat energy. Calculate the specific heat of this glass. • Givens: m = 4.0 gram q = 32 J ΔT = 40.0 K cp = ? J/g· K BE: q = mcΔT RE:

  6. A 4.0 gram sample of glass was heated from 274.0 K to 314.0 K and was found to absorb 32 J of heat energy. Calculate the specific heat of this glass. • Givens: m = 4.0 gram q = 32 J ΔT = 40.0 K cp = ? J/g· K BE: q = mcΔT q m x ΔT cp =

  7. 2. How much energy is needed to raise the temperature of a 1.60 gram sample of a metal from 273.0 K to 300.0 K, if the metal’s specific heat is 0.130 J/g ∙ K. • Givens: BE:

  8. How much energy is needed to raise the temperature of a 1.60 gram sample of a metal from 273.0 K to 300.0 K, if the metal’s specific heat is 0.130 J/g ∙ K. • Givens: m = 1.60 g q = ? ΔT = 7.0 K cp = 0.130 J/g· K BE: q = mcΔT q = (1.60 g) (0.130J/g ∙ K) (7.0 K) =

  9. 3. A 5.0 gram sample of a metal was heated from 25.0⁰ C to 40.0 ⁰C and it absorbed 17.6 J of energy. What is its specific heat? Identify the metal. • Givens: BE: RE:

  10. A 5.0 gram sample of a metal was heated from 25.0⁰ C to 40.0 ⁰C and it absorbed 17.6 J of energy. What is its specific heat? Identify the metal. • Givens: m = 5.0 g q = 17.6 J ΔT = 15 ⁰C cp = ? BE: q = mcΔT RE:

  11. 4. If 850.0 kJ of heat is transferred to 6000.0 grams of water at 280K, what will the final temperature of the water be? • Givens: m = q = ΔT = not needed Ti = Tf= ? cp = see page 60 BE: RE:

  12. Write the givens, basic equation and rearranged equation for the following problem. Check your units. If 850.0 kJ of heat is transferred to 6000.0 grams of water at 280.0 K, what will the final temperature of the water be?Givens: BE: RE:

  13. If 850.0 kJ of heat is transferred to 6000.0 grams of water at 280K, what will the final temperature of the water be? BE: q = mcΔT RE: Tf = + Ti • Givens: m = 6000.0 g q = 850000 J ΔT = not needed Ti = 280 K Tf= ? cp = 4.18 J/g· K q mc Tf = + Ti= + 280 = q 850000 J mc (6000 g) ( 4.18J/ g· K)

  14. FYI • Keep in mind that heat & specific heat are not the same thing. • Heat and energy are the same thing. • Cannot mix J and kJ in the same problem • Cannot mix g and kg in the same problem • To solve for final or initial temperature, substitute (Tf - Ti) for ΔT then rearrange equation to solve for desired variable. • Textbook problems will expect you to be able to apply previously taught material • You may need to refer to specific heat table on page 60

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