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 Specific Heat Capacity

Specific Heat Capacity (C or S ) - The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity of the substance. The quantity of heat is frequently measured in units of Joules(J). Another property, the specific heat, is the heat capacity of the substance per gram of the substance. The specific heat of water is 4.18 J/g° C.

 Substance C (J/g oC) Air 1.01 Aluminum 0.902 Copper 0.385 Gold 0.129 Iron 0.450 Mercury 0.140 NaCl 0.864 Ice 2.03 Water 4.18

q = m x C x DT

q = m x C x (Tf - Ti)

q = amount of heat energy gained or lost by substance

m = mass of sample

C = heat capacity (J oC-1 g-1 or J K-1 g-1)
Tf = final temperature
Ti = initial temperature

Specific Heat Instructional Videos

Solving for Heat

Solving for Mass
Solving for Specific Heat
Solving for Final Temperature

College Level Specific Heat Calculations

Solving for the Specific Heat of a Metal that is dropped in water.
 A 245.7g sample of metal at 75.2 degrees Celsius was placed in 115.43g water at 22.6 degrees Celsius. The final temperature of the water and metal was 34.6 Celsius. If no heat was lost to the surroundings what is the specific heat of the metal? Highlight Answer Below -qmetal=qwater  -(mCDT)=mCDT -(mC(Tf-Ti))=  mC(Tf-Ti) - (245.7g x C x (34.6oC-75.2oC))=  115.43g(4.18J/goC)(34.6oC-22.6oC) C x (9975goC)=5790J 0.580J/goC =C
 Solving for the Final Temperature when Metal is dropped in water. Determine the final temperature when a 25.0g piece of iron at 85.0°C is placed into 75.0grams of water at 20.0°C. The specific heat of iron is 0.450 J/g°C. The specific heat of water is 4.18 J/g°C. Highlight Answer Below -qmetal=qwater  -(mCDT)=mCDT -(mC(Tf-Ti))=  mC(Tf-Ti) -(25.0g(0.450J/goC)(Tf-85.0oC))=75.0g(4.18J/goC)(Tf-20.0oC) 956.25-11.25Tf=313.5Tf-6270 7226.25=324.75Tf 7226.25/324.75=Tf  22.3oC=Tf
Solving for Final Temperature when Ice is added to water
 What is the final temperature after a 21.5 gram piece of ice at 0 is placed into a Styrofoam cup with 125.0 grams of water initially at 76.5oC? Assume no loss or gain of heat from the surroundings. Highlight Answer Below energy to melt the ice                energy to bring the water to 0oC q=mHf                                vs.                          q=mCDT                        q=21.5 x 334j/g=7181J                                       q=125.0g (4.18J/goC)(76.5oC) =39,971J  The ice will melt, so the letft over energy is 39,971J  - 7181J= 32,790JReapply to q=mCDT, combine the mass of ice and water, assuming we are at a temp. of 0oC. 32,790J= 114.65g(4.18j/gC)(Tf-0) 32,790J/(114.65g(4.18j/gC))=Tf 53.5oC=Tf

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