1. Thermal Energy Transfer Laura Samide Andrew Weber

2. Calculating Thermal Energy Transfer The formula for Thermal Energy Transfer is q = nCpΔt 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 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. Energy transferred due to a temperature difference is heat. This is represented by q. q is the amount of heat gained or lost q is the amount of heat gained or lost The units for q is usually calories or joules 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 The quantity of the substance can be defined in either moles or grams This depends on the information that has been given This depends on the information that has been given Factor label can be used to convert from moles to grams and vice versa 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 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 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 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 The units of Cp depend on the other variables within the problem

6. Change in Temperature The change in temperature is represented by ΔT. 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, T i is higher than final temperature, ΔT = (T i – T f ) To find the change in temperature subtract the higher given temperature from the lower given temperature. i.e. if initial temperature, T i is higher than final temperature, ΔT = (T i – T f ) When a 200°C iron rod is put in a cold water bath the final temperature is 30°C When a 200°C iron rod is put in a cold water bath the final temperature is 30°C ΔT= 200 – 30= 170°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 (T i ) = 25.0ºC, and the final temp (T f ) = 30.0ºC Since you are trying to find the mass, isolate n first nCpΔt = Q n = Q / CpΔt Next substitute T f – T i for Δt n = Q / Cp(T f – T i )

8. Example 1 Continued Now you can plug in the values you were given for Q,Cp,T f, and T i n = (300J) / (4.19 J/ n = (300J) / (4.19 J/gºC)(30ºC – 25ºC) / (4.19 J/ n = (300J) / (4.19 J/gºC)(5ºC) / (20.95 J/g 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 (T i ) of 58ºC and a final temperature (T f ) of 82ºC with a specific heat (Cp) of 0.895 J/mºC. Find the heat lost or gained (q) of a 30 gram sample of aluminum at an initial temperature (T i ) of 58ºC and a final temperature (T f ) 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 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(T f -T i ) 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 = nCp. 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, T f: Q = nCp Q = nCpΔt (T f -T i ) Q = nCp(T f -T i ) Q/nCp = T f -T i (Q/nCp) + T i = T f

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

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