Thermodynamics and Entropy- reservoir and block problem

AI Thread Summary
A 382 g block is placed in contact with a thermal reservoir, initially at a lower temperature than the reservoir. The problem involves calculating the specific heat of the block using the change in entropy formula, ΔS = mCsp ln(T/To). The user initially misidentified the reference temperature To, leading to confusion in their calculations. After clarification, they realized the mistake and correctly identified the temperatures needed for the calculation. The specific heat of the block was ultimately calculated to be approximately 422.88 J/K*kg.
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?



NewPicture3.jpg

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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