Thermal Energy Thermal Energy TransferTransfer

Laura SamideLaura Samide

Andrew WeberAndrew Weber

Calculating Thermal Calculating Thermal Energy TransferEnergy Transfer

The formula for Thermal Energy The formula for Thermal Energy Transfer is q = nCpTransfer is q = nCpΔΔtt

Three of the variables will be Three of the variables will be given and you will have to solve given and you will have to solve for the other variablefor the other variable

HeatHeat

Energy transferred due to a Energy transferred due to a temperature difference is heat. This temperature difference is heat. This is represented by q.is represented by q.

q is the amount of heat gained or q is the amount of heat gained or lostlost

The units for q is usually calories or The units for q is usually calories or joulesjoules

Moles/GramsMoles/Grams

The quantity of the substance can The quantity of the substance can be defined in either moles or be defined in either moles or gramsgrams

This depends on the information This depends on the information that has been giventhat has been given

Factor label can be used to Factor label can be used to convert from moles to grams and convert from moles to grams and vice versavice versa

Specific HeatSpecific Heat

Specific heat is the amount of heat Specific heat is the amount of heat needed to raise the temperature of needed to raise the temperature of one gram of a substance one °C.one gram of a substance one °C.

Specific heat is represented by the Specific heat is represented by the variable Cpvariable Cp

The units for Cp are Joules/gram* °C, The units for Cp are Joules/gram* °C, Joules/mole* °C, calories/gram* °C, Joules/mole* °C, calories/gram* °C, and calories/mole* °Cand calories/mole* °C

The units of Cp depend on the other The units of Cp depend on the other variables within the problem variables within the problem

Change in Change in TemperatureTemperature The change in temperature is The change in temperature is

represented by represented by ΔΔT.T. To find the change in temperature To find the change in temperature

subtract the higher given temperature subtract the higher given temperature from the lower given temperature. i.e. if from the lower given temperature. i.e. if initial temperature, Tinitial temperature, Ti i is higher than final is higher than final temperature, temperature, ΔΔT =T = (T(Ti i – T– Tff) )

When a 200°C iron rod is put in a cold When a 200°C iron rod is put in a cold water bath the final temperature is 30°Cwater bath the final temperature is 30°C

ΔΔT= 200 – 30= 170°CT= 200 – 30= 170°C

Example 1Example 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 firstnCpΔt = Qn = Q / CpΔt

Next substitute Tf – Ti for Δt n = Q / Cp(Tf – Ti)

Example 1 ContinuedExample 1 Continued

Now you can plug in the values you Now you can plug in the values you were given for Q,Cp,Twere given for Q,Cp,Tff, and T, and Tii

n = (300J) / (4.19 J/n = (300J) / (4.19 J/gºC)(30ºC – 25ºC)n = (300J) / (4.19 J// (4.19 J/gºC)(5ºC)n = (300J) / (20.95 J/g / (20.95 J/g )n = 14.3 g

Example 2Example 2

Find the heat lost or gained (q) of a 30 gram Find the heat lost or gained (q) of a 30 gram sample of aluminum at an initial temperature sample of aluminum at an initial temperature (T(Tii) of 58ºC and a final temperature (T) of 58ºC and a final temperature (Tff) of ) of 82ºC with a specific heat (Cp) of 0.895 J/mºC.82ºC with a specific heat (Cp) of 0.895 J/mºC.

Since Cp uses moles as its units and not Since Cp uses moles as its units and not grams the mass of the aluminum will have to grams the mass of the aluminum will have to be converted from grams to molesbe converted from grams to moles

30g x 1m/27g =1.11m of aluminum30g x 1m/27g =1.11m of aluminum

Example 2 ContinuedExample 2 Continued

Now just plug in the numbers and solve for qNow just plug in the numbers and solve for q

q=nCp(Tq=nCp(Tff-T-Tii))

q=1.11m*0.895 J/mºC*(82ºC-58ºC)q=1.11m*0.895 J/mºC*(82ºC-58ºC)

q=1.11m*0.895 J/mºC*24ºCq=1.11m*0.895 J/mºC*24ºC

q=0.99345 J/ºC*24ºCq=0.99345 J/ºC*24ºC

q=23.8Jq=23.8J

Quiz 1Quiz 1

What is the heat change of the What is the heat change of the system if the mass is 20.0g, the system if the mass is 20.0g, the specific heat is 0.380 J/gºC and specific heat is 0.380 J/gºC and the change in temperature is the change in temperature is 25.0ºC25.0ºC

Quiz 1 SolutionQuiz 1 Solution

Since you are looking for heat (Q), Since you are looking for heat (Q), the variable is already isolated in the variable is already isolated in Q = nCpQ = nCpΔt..

Plug in the given values to find the Plug in the given values to find the answer:answer:

Q = 20g * 0.38J/gºC * 25ºCQ = 20g * 0.38J/gºC * 25ºC

Q = 190.0 JQ = 190.0 J

Quiz 2Quiz 2

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

Quiz 2 SolutionQuiz 2 Solution

First isolate your variable, TFirst isolate your variable, Tf:f:

Q = nCpQ = nCpΔt

Q = nCp(T(Tff-T-Tii))

Q/nCp = TQ/nCp = Tff-T-Tii

(Q/nCp) + T(Q/nCp) + Ti i = T= Tff

Quiz 2 SolutionQuiz 2 Solution

Now, plug in your given values:Now, plug in your given values:

TTf f = (1350J /70g * 2.12 J/gºC) + = (1350J /70g * 2.12 J/gºC) + 10.0ºC10.0ºC

TTf f = 19.1 ºC= 19.1 ºC

Works CitedWorks Cited

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