a) The molar heat capacity is 25.130 joules per gram per Kelvin.
b) The final temperature of the sample is 21.004 °C.
Procedure - Determination of the molar heat capacity and the final sensible temperature
a) Molar heat capacity of the sample
Dimensionally speaking, we can find the molar heat capacity ([tex]\bar c[/tex]), in joules per mol-Celsius, in terms of specific heat of iron ([tex]c[/tex]), in joules per kilogram-Celsius:
[tex]\bar c = c\cdot M[/tex] (1)
Where [tex]M[/tex] is the molar mass of iron, in grams per mole.
Please notice that [tex]\Delta T_{^{\circ}C} = \Delta T_{K}[/tex].
If we know that [tex]c = 0.45\,\frac{J}{g\cdot ^{\circ}C}[/tex] and [tex]M = 55.845\,\frac{g}{mol}[/tex], then the molar heat capacity:
[tex]\bar c = \left(0.45\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot \left(55.845\,\frac{g}{mol} \right)[/tex]
[tex]\bar c = 25.130\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The molar heat capacity is 25.130 joules per gram per Kelvin. [tex]\blacksquare[/tex]
b) Final temperature of the sample
The final temperature ([tex]T_{f}[/tex]), in degrees Celsius, is determined by definition of sensible heat ([tex]Q[/tex]), in joules:
[tex]T_{f} = T_{o}+\frac{Q}{n\cdot \bar {c}}[/tex] (2)
Where:
- [tex]T_{o}[/tex] - Initial temperature, in degrees Celsius.
- [tex]n[/tex] - Molar amount, in moles.
If we know that [tex]T_{o} = 12.2\,^{\circ}C[/tex], [tex]n = 5.6\,mol[/tex], [tex]\bar c = 25.130\,\frac{J}{g\cdot ^{\circ}C}[/tex] and [tex]Q = 1239\,J[/tex], then the final temperature of the sample is:
[tex]T_{f} = 12.2\,^{\circ}C + \frac{1239\,J}{(5.6\,mol)\cdot \left(25.130\,\frac{J}{g\cdot ^{\circ}C} \right)}[/tex]
[tex]T_{f} = 21.004\,^{\circ}C[/tex]
The final temperature of the sample is 21.004 °C. [tex]\blacksquare[/tex]
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