specific heat of material

Question

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

In the table, 𝑇1=45.7°C, 𝐸1=201J, 𝐸2=802J, 𝐸3=1470 J, and 𝐸4=2890J.

Other experiments determine that the material has a temperature of fusion of 𝑇fusion=239 °C and a temperature of vaporization of 𝑇vapor=483 °C.

If the sample of material has a mass of 𝑚=9.80 g, calculate the specific heat cs when this material is a solid. and when it is liquid, 𝑐l.

Answer 1

To calculate the specific heat when the material is in its solid state, we use the formula:

cs = (E2 - E1) / (mass × (Tfusion - T1))

Substituting the given values:

cs = (802 J - 201 J) / (9.80 g × (239 °C - 45.7 °C))

Now let's calculate the specific heat when the material is in its liquid state. We use the formula:

cl = (E4 - E3) / (mass × (Tvapor - Tfusion))

Substituting the given values:

cl = (2890 J - 1470 J) / (9.80 g × (483 °C - 239 °C))

Now you can plug in the values and calculate the specific heats. Let me know if you need further assistance!