Although the calorimeters are used to measure heat, the fact is that heat cannot be measured directly. Instead, the calorimeter is used to measure the change in temperature of a measured amount of a given substance. Heat is then calculated from this data; heat
is a derived quantity. Recall that the textbook defines specific heat capacity (c) as the quantity of heat required to raise the temperature of a gram of the substance 1  C. The equation is as follows. specific heat  mass quantity of
 temperatur heat e change
The equation used to calculate heat from experimental data can be obtained from the above equation simply by rearranging terms.
Multiplying both sides of the equation by (mass  temperature change) gives: q  mc  T where q  quanity of heat involved; m  mass; c  specific heat of substance;  T  change in temp.
To solve problems involving heat calculations, simply rearrange terms to isolate the unknown variable.
Example A
How much heat will be absorbed by 320 g of water when its temperature is raised by 35  C? The specific heat of water is 4.18 J/(g   C).
Solution q  mc  T q  q 
(320g)(4.1
8 J g   C
)( 35  C )
46 , 816 J  47,000J or 47 kJ (only 2 sig figs should be in answer)
Example B
Calculate the specific heat for aluminum if 16,500 J of heat are absorbed in raising the temperature of
1.50  10 2 g of aluminum by 125  C.
Solution q  mc  T
1 6 , 500 J  c 
(1.5
 10 2 g)(c )( 125  C )
0 .
880 J g   C
Now you try it!
1.
How much heat will be given off by 55 g of water as it cools from 87  C to 25  C?
ΔT  T f
 T i
 25  C  87  C  6 2  C q q q
 mc  T


(55g)(4.18
J g   C
)(  62  C)
 14,000J  14kJ (exothermi c)
2.
Calculate the specific heat of glass from the following data. The temperature of a piece of glass with a mass of 65 grams increases by 26  C when it absorbs 840 J of heat energy. q  mc  T
8 40 J c


(65g)(c)(2
0 .
50 J g   C
6  C)

3.
Calculate the temperature change for mercury if 160 grams of the metal absorb 1500 J of heat energy. Mercury’s specific heat is 0.14 J/(g   C). q
1500J


 T  mc  T
(160g)(0.1
4 J g   C
)(  T)
67  C
4.
The temperature of a 150 gram drinking glass decreased from 25  C to 14  C as 200 grams of water were added. Determine the original temperature of the water. (Hint: Heat loss must equal heat gain).
 q lost
 q gained
 q glass
 q water
 m water
 c water
 ΔT water
 (m glass
 c glass
 (150g)(0.5
0 J g  
 ΔT glass
C
)(  11  C)
 T water
)


(200g)(4.1
1  C
8 J g   C
)(  T water
)
 T
1  C
T i


T f

 14  C
13  C
T i
T i
5.
How much heat energy is required to raise the temperature of 200 grams of water from 25  C to
100  C? q  mc  T q  (200g)(4.1
8 J g   C
)( 7 5  C) q  62 , 700 J  60kJ (one sig.
fig.)
6.
The temperature of an iron bar with a mass of 87.0 g is raised from 31  C to 543  C. In the process
4.90  10 3 cal of heat energy were absorbed. What is the specific heat of iron?
ΔT  T f
 T i
 543  C  31  C  512  C (nearest whole #  3 sig figs) q  mc  T
4 .90
 10 3 cal c


(87.0g)(c) (512  C)
0 .
110 J g   C