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° 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?

Please can someone tell me how I would work this out step by step using the q=mcΔT formula.

Use the density and volume of the solid to compute the mass.

q = mass x specific heat x (Tfinal-Tinitial). solve for specific heat.

To find the specific heat of the solid in this experiment, you can use the formula q = mcΔT, where:

q = amount of energy (in Joules)
m = mass of the substance (in grams)
c = specific heat of the substance (in J/g°C)
ΔT = change in temperature (in °C)

In this case, you are provided with the following information:

Initial temperature (Ti) = 36.1°C
Final temperature (Tf) = 74.3°C
Volume of the solid (V) = 225.0 cm^3
Density of the solid at Ti = 1.75 g/cm^3
Energy (q) = 35850 J

To begin, we need to convert the volume of the solid to mass. Since density (d) is defined as mass (m) divided by volume (V), we can rearrange the formula to solve for mass:

m = d * V

Substituting the given values:
mass (m) = 1.75 g/cm^3 * 225.0 cm^3

Next, we can calculate the change in temperature (ΔT):

ΔT = Tf - Ti = 74.3°C - 36.1°C

Now we can substitute all the known values into the formula and solve for the specific heat (c):

q = mcΔT

Solving for c:

c = q / (m * ΔT)

Plug in the known values and calculate:

c = 35850 J / (mass * ΔT)

Just make sure all the units are coherent throughout the calculation.

After performing these calculations, you should obtain the specific heat of the solid.