Transcript Slide 1

Specific Heat Review
Cp = q/(m)(Δt)
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) = 8171.9 J
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
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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) = 3.636 cal/(gC)
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.
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Pick your equation
Your right!!!!...
q = Cp(m)(T) =
(2.09 J/gC)(34g)(67C - 55C) = 852.72J
Calculate the specific heat for a 102g sample that
requires 1430J to raise the temperature from 8.7C
to 12.5C.
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Pick your equation
Your right!!!..
Cp= q / (m)(T)
Cp= 1430J / (102g)(12.5C – 8.7C) = 3.69
J/(gC)
What will the final temperature be if, at room
temperature (25C), 1300 cal are added to a 76g
sample of iron?
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Pick your equation
Your right!!!........
Tf = [q /(m)(Cp)] + Ti
Tf= [1300cal / (76g)(0.11 cal/gC)] + 25C
= 180.5C
What would the change in temperature (T) be if
an 89g sample of copper required 678 calories of
heat?
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Pick your equation
Your right!!!!....
T = q / (m)(Cp) =
T = 678 cal/ (89g)(0.092 cal/gC) =
82.8C