A coffee cup calorimeter contains 25.0 grams water at 23.8 C
A 5.00g sample of an unknown metal at an initial temperature of 78.3 C was dropped into the calorimeter.
The final temperature of mixture was 46.3 C.
Calculate the specific heat of the metal. The specific heat of ater is 4.184 J/ (g C)?
You will need the following equation:
q=mc∆T
heat lost=heat gain, that is,
m1c1∆T1=m2c2∆T2
Where
m1=25.0g
c1=4.184 J/ (g C)
∆T1=46.3ºC-78.3º C=32.0ºC
m2=5.00g
c2=?
∆T2=46.3ºC-23.8ºC=22.5
Solve for c2,
c2=m1c1∆T1/m2∆T2
This must be a made up problem. I don't know that I've seen anything with a specific heat of this magnitude.
29.75
To calculate the specific heat of the metal, we can use the equation:
Q = m * c * ΔT
Where:
- Q is the heat transferred
- m is the mass of the metal (5.00g)
- c is the specific heat of the metal (what we need to find)
- ΔT is the change in temperature (final temperature - initial temperature)
First, let's calculate Q for the water in the calorimeter using the same equation:
Q_water = m_water * c_water * ΔT
Where:
- m_water is the mass of the water (25.0g)
- c_water is the specific heat of water (4.184 J/(g*C))
- ΔT is the change in temperature (final temperature - initial temperature)
Now, we know that the heat lost by the metal is equivalent to the heat gained by the water in the calorimeter:
Q_metal = -Q_water
Since the heat lost by the metal is equal to the heat gained by water, we can set up the following equation:
m_metal * c_metal * ΔT = -m_water * c_water * ΔT
Rearranging the equation, we get:
c_metal = (-m_water * c_water * ΔT) / (m_metal * ΔT)
To find the specific heat of the metal, substitute the given values into the equation:
c_metal = (-25.0g * 4.184 J/(g*C) * (46.3°C - 23.8°C)) / (5.00g * (78.3°C - 46.3°C))
By simplifying and calculating this equation, you'll find the specific heat of the metal.