The Specific heat of a substance is the energy (joules) required to raise one gram of substance by one degree celsius (units J/g °C).
Heating 225.0 cm3 of a solid from 36.1 °C to 74.3 °c takes 35850 J of energy. The density of the solid at 36.1 °C is 1.75 g/cm3.
What is the Specific Heat of the solid in this experiment
q = mass x specific heat x delta T.
Use the density to convert volume of the solid to mass. Solve for specific heat
Thaks would you be able to show me an examples, as I still don't quite understand it. Sorry but this is my first time doing chemistry.
To find the specific heat of the solid in this experiment, we can use the formula:
Q = mcΔT
Where:
Q is the energy transferred (35850 J),
m is the mass of the substance (solid),
c is the specific heat capacity of the substance (what we want to find),
ΔT is the change in temperature (74.3 °C - 36.1 °C = 38.2 °C).
First, we need to find the mass of the solid. We are given the volume of the solid (225.0 cm³) and its density at 36.1 °C (1.75 g/cm³).
The volume of the solid is given in cubic centimeters (cm³), and the density is given in grams per cubic centimeter (g/cm³). So, we can use the formula:
mass = density × volume
mass = 1.75 g/cm³ × 225.0 cm³
Now, let's calculate the mass of the solid.
mass = 1.75 × 225.0 = 393.75 grams
Now that we have the mass (393.75 grams), the energy transferred (35850 J), and the change in temperature (38.2 °C), we can substitute these values into the formula Q = mcΔT to solve for the specific heat capacity (c).
35850 J = (393.75 g) × c × (38.2 °C)
To isolate c, divide both sides of the equation by (393.75 g) × (38.2 °C):
c = (35850 J) / [(393.75 g) × (38.2 °C)]
Calculate the value of c using the given values:
c = 35850 J / (393.75 g × 38.2 °C)
c ≈ 0.243 J/g °C
Therefore, the specific heat of the solid in this experiment is approximately 0.243 J/g °C.