an experiment is performed on an unknown material and produces the following hea

Question

an experiment is performed on an unknown material and produces the following heat curve that temperature of that material is shown as a function of heat added if the sample of the material has a mass of m equals 8.00 grams calculate the specific heat when this material is a solid and when it is a liquid (please see the hidden panel for they split temperatures of fusion and vaporization do not try to approximate the graph). hint: this material is found to have a temperature of fusion of tea Fusion equals 237 Celsius and a temperature of vaporization of T Vapor equals 479 Celsius use this information along with the points of interest given on the graph in the heat equation to find the specific heat of the material.

iscms/mod/ibis/view.php?id-4246689 ng Learning experiment is performed on an unknown material and produces the following heat curve. The mperature of the material is shown as a function of heat added. If the sample of material has a mass of 8.00 g, calculate the specific heat when this material is a solid, cs, and when it is liquid, q. (Please se e hint panel for the explicit temperatures of fusion and vaporization, do not try to approximate from the aph.) Temperature (Celsius) 550 500 450 400 350 300 250 200 150 100 45.20 100 175 785 1410 2810 Heat (J) Number Number O Previous Check Answer 0 Next terma of use

Explanation / Answer

Cs = specific heat of solid

Q = m*C*(Tf-Ti)

if m = 8 g then

slope = Change in heat / (change in T)

slope = (785-175)/(237-45.2)

slope = 3.3926 J/°C

yet, we need per unit mass

3.3926 J/gc * 1/(8g) = 0.39754 J/gC

now...

Cl = liquid

choose

slope2 = Q2/(Tf-Ti) = (2810-1410)/(479-237)

slop2 = 5.490 J/gC

per unit mas

5.490/8 = 0.72314J/gC for liquid

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