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Transcript
UNIT 3: Energy 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 Air Aluminum Copper Gold Iron Mercury NaCl Ice Water C (J/g oC) 1.01 0.902 0.385 0.129 0.450 0.140 0.864 2.03 4.18 q = m x C x T 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 Solving for Specific Heat of a metal 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? -qmetal=qwater -(mCT)=mCT -(mC(Tf-Ti))= mC(Tf-Ti) o - (245.7g x C x (34.6 C-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. -qmetal=qwater -(mCT)=mCT -(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. energy to melt the ice q=mHf q=21.5 x 334j/g=7181J energy to bring the water to 0oC vs. q=mCT q=125.0g (4.18J/goC)(76.5oC) =39,971J The ice will melt, so the letft over energy is 39,971J - 7181J= 32,790J Reapply to q=mCT, combine the mass of ice and water, assuming we are at a temp. of 0 oC. 32,790J= 1146.5g(4.18j/gC)(Tf-0) 32,790J/(1146.5g(4.18j/gC))=Tf 53.5oC=Tf