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Thermal Energy Transfer

Thermal Energy Transfer. Laura Samide Andrew Weber. Calculating Thermal Energy Transfer. The formula for Thermal Energy Transfer is q = nCp Δ t Three of the variables will be given and you will have to solve for the other variable. Heat.

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Thermal Energy Transfer

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  1. Thermal Energy Transfer Laura Samide Andrew Weber

  2. Calculating Thermal Energy Transfer • The formula for Thermal Energy Transfer is q = nCpΔt • Three of the variables will be given and you will have to solve for the other variable

  3. Heat • Energy transferred due to a temperature difference is heat. This is represented by q. • q is the amount of heat gained or lost • The units for q is usually calories or joules

  4. Moles/Grams • The quantity of the substance can be defined in either moles or grams • This depends on the information that has been given • Factor label can be used to convert from moles to grams and vice versa

  5. Specific Heat • Specific heat is the amount of heat needed to raise the temperature of one gram of a substance one °C. • Specific heat is represented by the variable Cp • The units for Cp are Joules/gram* °C, Joules/mole* °C, calories/gram* °C, and calories/mole* °C • The units of Cp depend on the other variables within the problem

  6. Change in Temperature • The change in temperature is represented by ΔT. • To find the change in temperature subtract the higher given temperature from the lower given temperature. i.e. if initial temperature, Ti is higher than final temperature, ΔT =(Ti – Tf) • When a 200°C iron rod is put in a cold water bath the final temperature is 30°C ΔT= 200 – 30= 170°C

  7. Example 1 Find the mass of the substance if Cp = 4.19 J/gºC, Q = 300.0J, the initial temp (Ti) = 25.0ºC, and the final temp (Tf) = 30.0ºC Since you are trying to find the mass, isolate n first nCpΔt = Q n = Q / CpΔt Next substitute Tf – Ti for Δt n = Q / Cp(Tf – Ti)

  8. Example 1 Continued Now you can plug in the values you were given for Q,Cp,Tf, and Ti n = (300J) / (4.19 J/gºC)(30ºC – 25ºC) n = (300J) / (4.19 J/gºC)(5ºC) n = (300J) / (20.95 J/g ) n = 14.3 g

  9. Example 2 • Find the heat lost or gained (q) of a 30 gram sample of aluminum at an initial temperature (Ti) of 58ºC and a final temperature (Tf) of 82ºC with a specific heat (Cp) of 0.895 J/mºC. • Since Cp uses moles as its units and not grams the mass of the aluminum will have to be converted from grams to moles 30g x 1m/27g =1.11m of aluminum

  10. Example 2 Continued Now just plug in the numbers and solve for q q=nCp(Tf-Ti) q=1.11m*0.895 J/mºC*(82ºC-58ºC) q=1.11m*0.895 J/mºC*24ºC q=0.99345 J/ºC*24ºC q=23.8J

  11. Quiz 1 What is the heat change of the system if the mass is 20.0g, the specific heat is 0.380 J/gºC and the change in temperature is 25.0ºC

  12. Quiz 1 Solution Since you are looking for heat (Q), the variable is already isolated in Q = nCpΔt. Plug in the given values to find the answer: Q = 20g * 0.38J/gºC * 25ºC Q = 190.0 J

  13. Quiz 2 Find the final temperature if 1350J of heat is added to a 70g system with an initial temperature of 10.0ºC and a specific heat of 2.12 J/gºC

  14. Quiz 2 Solution First isolate your variable, Tf: Q = nCpΔt Q = nCp(Tf-Ti) Q/nCp = Tf-Ti (Q/nCp) + Ti = Tf

  15. Quiz 2 Solution Now, plug in your given values: Tf = (1350J /70g * 2.12 J/gºC) + 10.0ºC Tf = 19.1 ºC

  16. Works Cited • Smoot, Robert, Jack Price, and Richard Smith. Chemistry a Modern Course. Columbus: Merrill Publishing Company, 1987.

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