1. Specific Heat Review
Cp = q/(m)(Δt)

2. Calculate the amount of heat required to raise the temperature of 78
Calculate the amount of heat required to raise the temperature of 78.2g of water from 10C to 35C.
Manipulate the formula to solve for heat:
Cp= q solve for q = (Cp)(m)(T)
m Δt
Check the units: m = 78.2g, T1 = 10C, T2 = 35C
Cp is not given, but we are dealing with water so Cp = 4.18 J/(gC)
Heat = ? J
All the units match-up, so we do not have to convert any units prior to plugging into the equation.
Heat = Cp(m)(T) Heat = [4.18 J/(gC)] (78.2g)(35C - 10C) = J

3. Manipulate the formula to solve for specific heat: Cp = q/ (m)(T)
What is the specific heat of a 75 gram sample that requires 1200cal to change the temperature from 25C to 85F?
Manipulate the formula to solve for specific heat: Cp = q/ (m)(T)
Check the units: m = 75g, Heat (q) = 1200cal,
T1 = 25C, T2 = 85F
Cp = ? cal/(gC)
T2 needs to be converted from F to C: C = 5/9 (F – 32) => 5/9 (85 – 32) = 29.4C = T2
Cp = q / (m)(T) Cp = 1200cal / (75g)(29.4C - 25C) = cal/(gC)

4.   Calculate the amount of heat, in joules, need to raise 34g of ice (Cpice= 2.09 J/gC) from 55C to 67C.  
Pick your equation
Your right!!!!...
q = Cp(m)(T) =
(2.09 J/gC)(34g)(67C - 55C) = J

5. Cp= 1430J / (102g)(12.5C – 8.7C) = 3.69 J/(gC)
Calculate the specific heat for a 102g sample that requires 1430J to raise the temperature from 8.7C to 12.5C.
Pick your equation
Your right!!!..
Cp= q / (m)(T)
Cp= 1430J / (102g)(12.5C – 8.7C) = 3.69 J/(gC)

6. Tf= [1300cal / (76g)(0.11 cal/gC)] + 25C = 180.5C
What will the final temperature be if, at room temperature (25C), 1300 cal are added to a 76g sample of iron?  
Pick your equation
Your right!!!
Tf = [q /(m)(Cp)] + Ti
Tf= [1300cal / (76g)(0.11 cal/gC)] + 25C = 180.5C

7. T = 678 cal/ (89g)(0.092 cal/gC) = 82.8C
What would the change in temperature (T) be if an 89g sample of copper required 678 calories of heat?
Pick your equation
Your right!!!!....
T = q / (m)(Cp) =
T = 678 cal/ (89g)(0.092 cal/gC) = 82.8C