Chemical Systems & Heat

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Presentation transcript:

Chemical Systems & Heat Unit 10 Review

Calculate the final temp 14.0 g of metal at 24.0 C has 250 joules of heat added to it. The metal’s specific heat is 0.105 J/gC. What is its final temperature? q= mc ΔT q = m c TF – TI TF = 250 J / (14.0g)(0.105 J/g ⁰C) + 24 ⁰C = 28 ⁰C

What will the final temperature be? 50.0 g iron with an initial temperature of 225⁰C and 50.0g of gold with an initial temperature of 25.0⁰C are brought into contact with one another. No heat is lost to the surroudings. What will the temperature be when the two metals reach thermal equilibrium? Specific heat iron = 0.449 J/g⁰C Specific heat gold = 0.128 J/g⁰C

Specific heat = ? 100.0 g of nickel @ 150⁰C was placed in 1.00 L of water at 25.0 ⁰C. The final temperature of the water was 26.3⁰C. What is the specific heat of nickel? 0.44 J/g⁰C

Final temperature of the water? A 25.0 g piece of iron @ 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water @ 298 K. No heat is lost to the cup or the surroundings. What will the final temperature of the water be? Specific heat of iron = 0.499 J/g⁰C 34.6 ⁰C