You heat 22.05 g of a solid in a test tube to 100.00°C and add it
to 50.00 g of water in a coffee-cup calorimeter. The water
temperature changes from 25.10°C to 28.49°C. Find the specific
heat capacity of the solid.
To find the specific heat capacity of the solid, we can use the equation:
Q = m * c * ΔT
Where Q is the heat absorbed or released by the substance, m is its mass, c is its specific heat capacity, and ΔT is the change in temperature.
In this case, we know the following values:
M (mass of the solid) = 22.05 g
M (mass of water) = 50.00 g
T initial (initial temperature of water) = 25.10°C
T final (final temperature of water) = 28.49°C
First, let's calculate the heat absorbed by the water. We can use the equation:
Q water = m * c * ΔT
Q water = 50.00 g * 4.18 J/g°C * (28.49°C - 25.10°C)
Q water = 50.00 g * 4.18 J/g°C * 3.39°C
Q water = 709.77 J
Next, let's calculate the heat released by the solid. Since the heat released by the solid is equal to the heat absorbed by the water (assuming no heat is lost to the surroundings), we can use the same value:
Q solid = 709.77 J
Lastly, let's substitute the values into the equation for the solid:
Q solid = m * c * ΔT
709.77 J = 22.05 g * c * (100.00°C - T initial)
Solving for c:
c = Q solid / (m * ΔT)
c = 709.77 J / (22.05 g * (100.00°C - 25.10°C))
c ≈ 0.324 J/g°C
Therefore, the specific heat capacity of the solid is approximately 0.324 J/g°C.