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Chemical Systems & Heat

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 Δ T q = m c T F – T I T F = 250 J / (14.0g)(0.105 J/g ⁰C) + 24 ⁰C = 28 ⁰C.

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Chemical Systems & Heat

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  1. Chemical Systems & Heat Unit 10 Review

  2. 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

  3. 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 25.0⁰Care 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

  4. 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

  5. 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

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