Entrance Ticket Part I 3/10/15
Add the following to 17.1 #5-9
1.) How much energy in Joules is
needed to heat up 15.0 grams of
water from 7.25 ºC to 38.0 ºC?
(specific heat of water is 4.184 J/gºC)
Entrance Ticket Part II 3/10/15
2.) When a certain metal is cooled from
130 ºC to 22 ºC it releases 415 Joules.
What is the mass of the metal if its
specific heat capacity is 0.24 J/gºC.
3.) How much energy is needed to
melt a 100. g chunk of ice at 0 ºC?
(heat of fusion is 334 J/g)
Answers
2.)
3.)
Q = mcΔT
415 J = m x 0.24 J/gºC x 108 ºC
m = 16 g
Q = m ΔHfus
Q = 100. g x 334 J/g
Q = 33,400 J
Entrance Ticket 3/12/15
1.) 12,000 Joules of energy are
needed to heat up a certain mass of
water by 10. ºC. What is the mass of
water?
(specific heat of water is 4.18 J/gºC)
Q = mcΔT
12,000 J = m x 4.18 J/gºC x 10. ºC
m = 290 g
Specific Heat Practice
1.) 24,000 J or 5,700 cal
2.) 8.9 x 106 J or 8,900,000 J
3.) 97 ºC
4.) 36.2 g
5.) 1,200 cal
Specific Heat Practice
6.) 0.838 J/gºC
7.) 1,690 g
8.) 2,980 kJ
9.) 3.0 ºC
10.) 0.89 J/gºC Aluminum
Heat Problems Cont…
How many Joules would it take to heat
250. g of ice from 0ºC to 100 ºC?
(specific heat of water is 4.184 J/g ºC &
the heat of fusion is 334 J/g)
Phase Δ & Temp. Δ
Q = m ΔHfus
Q = 250g x 334 J/g
Q = 83500 J
Q = mcΔT
Q = 250g x 4.184 J/gºC x 100ºC
Q = 104600 J
Q = 83500 + 104600 = 188,100
J
Heat Problems Cont…Honors
How many Joules would it take to heat
125 g of ice from -10ºC to 95ºC?
-specific heat of ice is 2.1 J/gºC,
-specific heat of water is 4.184 J/gºC
-the heat of fusion of H20 is 334 J/g
-the freezing point of water is 0ºC
Answer
Q=125g x 2.1 J/gºC x 10ºC = 2,625 J
Q= 125g x 334 J/g = 41,750 J
Q=125g x 4.184 J/gºC x 95ºC = 49,685 J
Total Q = 94,000 J
-specific heat of ice is 2.1 J/gºC,
-specific heat of water is 4.184 J/gºC
-the heat of fusion of H20 is 334 J/g
-the freezing point of water is 0ºC
Heat Problems Cont…Honors
How many Joules would it take to
melt 65 g of ice that has a temp. of
-7.5ºC?
-specific heat of ice is 2.1 J/gºC,
-the heat of fusion of H20 is 334 J/g
-the freezing point of water is 0ºC
Q = mcΔT
Q = 65 g x 2.1 J/gºC x 7.5 ºC
Q =1024 J
Q = mΔHfus
Q = 65 g x 334 J/g
Q = 21,710 J
1024 J + 21,710 J = 23,000 J
17 Quiz Friday
-Know what heat is.
-Know how heat flows, and how it is related to
kinetics.
-Know how to solve phase change & temperature
change problems involving heat transfer.
-Know what endothermic and exothermic mean.
-Know what latent heat is, and when it’s used.
-Know what specific heat capacity is, and when
it’s used.
-1 Cal = 1000 cal
-Understand your labs
150
Cheeto
Lab
p. 33
www.fritolay.com
Cheeto Lab
Objective:
Determine the number of Calories per gram
in a Cheeto.
Materials:
Soda can calorimeter (measures heat)
Cheeto
Pins
Graduated cylinder
Mass scale
Water
Thermometer
Matches
Cheeto Lab
Procedure:
Write your own procedure!
Hints:
Q = mcΔT
cwater = 1 cal/g°C
1000 cal = 1 Calorie
Assumption: The energy gained by the water
will be equal to the energy lost by the Cheeto!
Analysis:
1. Show your data and calculations.
a. Calculate the heat transferred (the
calories of your Cheeto)
b. Convert to food Calories
c. Divide by the mass of the Cheeto
d. Compare to the food label
2. Calculate the % yield. (Standard 3f *)
 actual. yield

 theoretical. yield

  100  percent. yield

Cheeto Lab
Grading:
Every lab partner must have their own
identical lab. Turn in one lab per group.
All the labs must be complete before the
one sample is turned in!
Failure to create your copy of the lab will
result in a zero on your lab grade.
Cheeto Lab
Grading:
Procedure
Analysis
Safety
Total
5 points
15 points (3 each)
5 points
25 points
© 2007 BPHS Mr. Reule / Mr. Melton