Thermodynamics and Entropy- reservoir and block problem

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SUMMARY

The discussion centers on calculating the specific heat capacity (Csp) of a 382 g block in contact with a thermal reservoir, where the block initially has a lower temperature than the reservoir. The entropy change (ΔS) is given as 100 J/K, and the temperatures involved are Ta = 280 K and Tb = 520 K. The correct formula for specific heat capacity is derived as Csp = ΔS / (m * ln(T / To)), leading to a calculated value of 422.88 J/K·kg. A critical error identified was the incorrect assignment of the initial temperature (To) in the calculation.

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emmy
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Homework Statement


A 382 g block is put in contact with a thermal reservoir. The block is initially at a lower temperature than the reservoir. Assume that the consequent transfer of energy as heat from the reservoir to the block is reversible. Figure 20-22 gives the change in entropy ΔS of the block until thermal equilibrium is reached. The scale of the horizontal axis is set by Ta = 280 K and Tb = 520 K. What is the specific heat of the block?



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Homework Equations


ΔS= mCspln(T/To)

The Attempt at a Solution


ΔS= mCspln(T/To)

Csp=ΔS/(m*ln(T/To))

where
ΔS= 100 J/K
m= .382 kg
T=520K
To=280K

plug and chug,
Csp= 100/(.382*ln(520/280)) =422.88J/K*kg

where did I go wrong ):
 
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your T0 is not Ta

look carefully at the graph^^
 
quietrain said:
your T0 is not Ta

look carefully at the graph^^


Ohhhhh my gosh this is why i miss points on exams lol! Thanks c:
 

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