A 75.00-g sample of an unknown metal is heated to 52.00°C. It is then placed in a coffee cup calorimeter filled with water. The calorimeter and the water inside have a combined mass of 225.0 g and an overall specific heat of 2.045 J/(gi°C). The initial temperature of the calorimeter is 7.50°C. If the system reaches a final temperature of 9.08°C when the metal is added, what is the specific heat of the unknown metal?
given:
m1 = 75.00 g
c1 = ?
t1 = 52.00°C
m2=225.0 g
c2=2.045 J/g-°C
t2=07.50°C
tf=9.08°C
Using
ΣmcΔt=0
75.00*c1*(9.08-52.00)+225.0(2.045)(9.08-7.50)=0
Solve for c1 to get c1=0.2258 J/g-°C
To find the specific heat of the unknown metal, we can use the principle of energy conservation.
First, let's calculate the heat transferred to or from the water and calorimeter using the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the water and calorimeter (225.0 g)
c is the specific heat of the water and calorimeter (2.045 J/(g•°C))
ΔT is the change in temperature (final temperature - initial temperature)
Calculating Q, we have:
Q = (225.0 g) * (2.045 J/(g•°C)) * (9.08°C - 7.50°C)
Q = 775.65 J
Since energy is conserved, the heat transferred to or from the water and calorimeter is equal to the heat transferred to or from the metal. So we can write:
Q_metal = -Q
Now, we can find the heat transferred to or from the metal using the formula:
Q_metal = m_metal * c_metal * ΔT
Where:
Q_metal is the heat transferred to or from the metal, which is equal to -Q
m_metal is the mass of the metal (75.00 g)
c_metal is the specific heat of the metal (unknown)
ΔT is the change in temperature (final temperature - initial temperature)
Plugging in the values, we have:
-775.65 J = (75.00 g) * c_metal * (9.08°C - 52.00°C)
Simplifying the equation:
-775.65 J = -29214 g•°C * c_metal
Dividing both sides by -29214 g•°C:
c_metal = -775.65 J / -29214 g•°C
c_metal ≈ 0.0266 J/(g•°C)
Therefore, the specific heat of the unknown metal is approximately 0.0266 J/(g•°C).