Answer:
An irregular lump of an unknown metal has a measured density of 2.97 g/mL. The metal is heated to a temperature of 173 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL, and the temperature is recorded as 40.6 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.
Chemistry
2 Answers
glerp · Truong-Son N.
Feb 4, 2018
c
=
0.46
J
g
⋅
°
C
Explanation:
Write down given information for both substances (mass, specific heat, change in temperature)
For Water :
m
=
25.0
g
c
=
4.181
J
g
⋅
°
C
T
i
n
i
t
i
a
l
=
25
°
C
T
f
i
n
a
l
=
40.6
°
C
Δ
T
=
T
f
i
n
a
l
−
T
i
n
i
t
i
a
l
=
40.6
°
C
−
25
°
C
=
15.6
°
C
For Metal :
m
=
D
⋅
V
=
2.97
g
m
L
⋅
9
m
L
=
26.73
g
c
=
?
T
i
n
i
t
i
a
l
=
173
°
C
T
f
i
n
a
l
=
40.6
°
C
Δ
T
=
T
f
i
n
a
l
−
T
i
n
i
t
i
a
l
=
40.6
°
C
−
173
°
C
=
−
132.4
°
C
Use the formula
Q
=
m
c
Δ
T
to find the change in energy in water.
Q
w
=
(
25.0
g
)
(
4.181
J
g
⋅
°
C
)
(
15.6
°
C
)
Q
w
=
1630.59
J
This (
Q
w
) is the amount of energy the water gained, so this means the metal lost the same amount(
−
Q
m
), according to the law of conservation of energy.
Q
w
=
−
Q
m
Rearrange the formula
Q
=
m
c
Δ
T
to find
c
of the metal.
c
=
Q
m
m
⋅
Δ
T
c
=
−
1630.59
J
(
26.73
g
)
(
−
132.4
°
C
)
c
=
0.46
J
g
⋅
°
C
Explanation:
hope s hlp
