Chemical Systems & Heat

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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 – T I
• 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

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