A piece of metal of mass 25 g at 94◦C is placed
in a calorimeter containing 57.4 g of water at
24◦C. The final temperature of the mixture
is 34.2
◦C. What is the specific heat capacity
of the metal? Assume that there is no energy
lost to the surroundings.
Answer in units of J
g ·
◦ C
.
To find the specific heat capacity of the metal, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat transferred (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g · ◦C)
ΔT = change in temperature (in ◦C)
In this case, we need to find the specific heat capacity of the metal, so we'll call it "cm".
First, let's find the heat transferred to the metal:
Qm = cm * mm * ΔTm
Where:
Qm = heat transferred to the metal
mm = mass of the metal (25 g)
ΔTm = change in temperature of the metal (final temperature - initial temperature)
To find ΔTm, we subtract the initial temperature of the metal from the final temperature:
ΔTm = 34.2◦C - 94◦C
Now, let's find the heat transferred to the water:
Qw = cw * mw * ΔTw
Where:
Qw = heat transferred to the water
mw = mass of the water (57.4 g)
cw = specific heat capacity of water (4.18 J/g · ◦C)
ΔTw = change in temperature of the water (final temperature - initial temperature of the water)
To find ΔTw, we subtract the initial temperature of the water from the final temperature:
ΔTw = 34.2◦C - 24◦C
Since there is no energy lost to the surroundings, the heat transferred to the metal is equal to the heat transferred to the water:
Qm = Qw
cm * mm * ΔTm = cw * mw * ΔTw
Now, we can solve for cm:
cm = (cw * mw * ΔTw) / (mm * ΔTm)
Plugging in the given values:
cm = (4.18 J/g · ◦C * 57.4 g * (34.2◦C - 24◦C)) / (25 g * (34.2◦C - 94◦C))
Simplifying:
cm = (4.18 J/g · ◦C * 57.4 g * 10.2◦C) / (25 g * -59.8◦C)
cm ≈ - 16.5928 J/g · ◦C
Therefore, the specific heat capacity of the metal is approximately -16.5928 J/g · ◦C.