The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:
[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)
Where:
- [tex]\Delta T[/tex] - Temperature change, in degrees Celsius.
- [tex]\Delta Q[/tex] - Heat received, in joules.
- [tex]m[/tex] - Mass of the sample, in grams.
- [tex]c[/tex] - Specific heat of the sample, in joules per kilogram-degrees Celsius.
Solid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]
[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
Liquid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]
[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
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