A 19.5-g sample of a metal was heated to 61.67°C. When the metal was placed into 27.1 g of water in a calorimeter, the temperature of the water increased from 25.00°C to 30.00°C. What is the specific heat of the metal?
To find the specific heat of the metal, we can use the formula:
q = m * c * ΔT
Where:
- q is the heat absorbed or released
- m is the mass
- c is the specific heat
- ΔT is the change in temperature
In this case, the heat absorbed by the water can be calculated using the formula:
q_water = m_water * c_water * ΔT_water
We are given the following values:
- m_water = 27.1 g (mass of water)
- c_water is the specific heat of water, which is considered to be 4.18 J/g°C.
- ΔT_water = 30.00°C - 25.00°C = 5.00°C
Substituting the known values into the equation, we have:
q_water = 27.1 g * 4.18 J/g°C * 5.00°C
Simplifying the equation:
q_water = 567.53 J
Now, let's calculate the heat released or absorbed by the metal. Since the metal was heated and released heat, the equation becomes:
q_metal = -q_water
Therefore:
q_metal = -567.53 J
Now, we can substitute the known values for the mass and the change in temperature of the metal into the equation:
q_metal = 19.5 g * c_metal * (61.67°C - 30.00°C)
Simplifying the equation:
-567.53 J = 19.5 g * c_metal * 31.67°C
To find the specific heat of the metal, we can rearrange the equation:
c_metal = -567.53 J / (19.5 g * 31.67°C)
Calculating this expression:
c_metal ≈ -0.936 J/g°C
Therefore, the specific heat of the metal is approximately -0.936 J/g°C.