An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as a function of heat added.

Question

In the table, 𝑇1=48.2 °C, 𝐸1=173 J, 𝐸2=819 J, 𝐸3=1430 J, and 𝐸4=2870 J.

Other experiments determine that the material has a temperature of fusion of 𝑇fusion=235 °C and a temperature of vaporization of 𝑇vapor=483 °C.

If the sample of material has a mass of 𝑚=9.20 g, calculate the specific heat when this material is a solid, 𝑐s, and when it is liquid, 𝑐l.

Answer 1

Hi Ann, I can give this one a try.

First let's write out what the problem gives us:

𝑇1=48.2 °C, 𝐸1=173 J, 𝐸2=819 J, 𝐸3=1430 J, and 𝐸4=2870 J, 𝑇fusion=235 °C, 𝑇vapor=483 °C, m = 9.20 g.

We need to find: Cs and Cl.

To find Cs we need to use the formula for "MCAT":

Q = Change in E = m*Cs*(Change in T), where Change in E = E2 - E1 = (819 - 173) J = 646 J. m = 9.20 g. Change in T is Tf - T1 = (235 - 48.2)°C = 186.8°C.

Using these values for equation Q = m*Cs*(Change in T), we get:

646 J = (9.20 g)*(Cs)*(186.8°C) => Cs = 646 J/1718.56 g*°C = 0.376 J/g*°C.

Cs = 0.376 J/g*°C.

For Cl:

We need to use the same formula Q = Change in E = m*Cv*(Change in T). Notice how now we have Cv instead of Cs. Change in E = E4 - E3 = (2870 - 1430) J = 1440 J, Change in T = Tv - Tf = (483 - 235)°C = 248°C, m = 9.20 g.

Plugging these numbers into our equation, we get:

(1440 J) = (9.20 g)*(Cv)*(248°C) => Cv = 1440 J/2281.6 g*°C= 0.631 J/g*°C.

Cl = 0.631 J/g*°C.

I hope that helps! Please let me know if you have any questions. All the best!