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 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 25.0⁰C 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 = J/g⁰C
Specific heat gold = J/g⁰C

4. Specific heat = ?
100.0 g of 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 398 K is placed in a styrofoam coffee cup containing 25.0 mL of 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 = J/g⁰C
34.6 ⁰C