Question

An experiment is performed on an unknown material and produces the following heat curve. The temperature of the material is shown as a function of heat added. 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, q. (Please see the hint panel for the explicit temperatures of fusion and vaporization, do not try to approximate from the graph.) Temperature (Celsius) 550 500 450 400 350 300 250 200 150 100 47.250 -50 100 802 1510 2850 Heat (J) Number Number

Answer 1

first line is line when material is in solid state and 3rd line when material exist only in liquid state

Now slope of first line = Change in heat / (change in T)

slope = (802-171)/(230-47.2)

slope = 3.45 J/°C

yet, we need per unit mass

3.45 J/^oc * 1/(9.2g) = 0.375 J/g^oC = C_s

again, slope of second line = Change in heat / (change in T)

slope = (2850-1510) / (485-230) = 5.25 J/^oc

So, C_l = 5.25 J/^oc / 9.20 g = 0.571 J/g.^oC