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Moderns that are heating up on ebay!
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Started by thirdgreenham,

63,755 posts in this topic

gradejunky

gradejunky

Posted
Posted
On 2/10/2023 at 2:48 PM, PeterPark said:

Whatdya wanna know?

Oh, nothing too difficult.  Although earlier today, I was pondering this as I walked to the mailbox:

Suppose a piece of solid bismuth weighing 27.7 g at a temperature of 253 °C is placed in 277 g of liquid bismuth at a temperature of 333 °C. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings. The enthalpy of fusion of solid bismuth is ΔHfus = 11.0 kJ mol–1 at its melting point of 271 °C, and the molar heat capacities CP of solid and liquid bismuth are 26.3 and 31.6 J K–1 mol–1, respectively.

 

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Jesse-Lee

Jesse-Lee

Posted
Posted
On 2/10/2023 at 2:33 PM, gradejunky said:

Oh, nothing too difficult.  Although earlier today, I was pondering this as I walked to the mailbox:

Suppose a piece of solid bismuth weighing 27.7 g at a temperature of 253 °C is placed in 277 g of liquid bismuth at a temperature of 333 °C. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings. The enthalpy of fusion of solid bismuth is ΔHfus = 11.0 kJ mol–1 at its melting point of 271 °C, and the molar heat capacities CP of solid and liquid bismuth are 26.3 and 31.6 J K–1 mol–1, respectively.

I'm no scientist, but I'd guess it'd probably be 401 degrees Celsius.

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the blob

the blob

Posted
Posted (edited)
On 2/9/2023 at 1:27 PM, OtherEric said:

Actually, I may have misspoke... it looks like lots of the recent sales are for copies under 9.8 or from other companies, so it probably is at a new peak.  I was thinking they had hit around $1000 then dropped back to the $700 range, but I haven't looked all that often recently.  My copy has a small color breaking corner tic on the back, so I haven't bothered getting it graded.

Supergirl_Legion_23_var.thumb.jpg.a8d7682ad1871df7e228d9fbda73d8d5.jpg

Need to get my copy slabbed. Possibly my best dollar box find ever. Possibly 15 for $10.

 

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Jesse-Lee

Jesse-Lee

Posted
Posted
On 2/10/2023 at 5:13 PM, GeeksAreMyPeeps said:

I recently picked up a handful of those from Midtown Comics; looks like they're sold out now.

I snagged a couple at cover price as well - I looked at a few places online and it looks like they're sold out (at least where I was looking). There's one seller asking $20 on eBay that I could find, and a few Canadian sellers with copies from about $6-10, but with $10-17 shipping...

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Corona smith

Corona smith

Posted
Posted (edited)
On 2/1/2023 at 2:44 PM, godzilla43 said:

I don't have 655,656 or 657 but in 655 Damian is only seen in shadows? 656 in last page he attacks Batman and he speaks? 657 he is in the cover and in the book, talks and he is named? 
So ain't 657 then Damians first appearance? Like ASM 299 is cameo and Hulk 180 is cameo.

Batman #656 is a first full appearance regardless of what anyone states here. Damian appears in 5 panels which to me is a first full. 

Edited by Corona smith
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Lazyboy

Lazyboy

Posted
Posted
On 2/11/2023 at 7:15 AM, Corona smith said:

Batman #656 is a first full appearance regardless of what anyone states here.

No, regardless of what anyone states here..

On 2/11/2023 at 7:15 AM, Corona smith said:

Damian appears in 5 panels which to me is a first full. 

It's not, and the number of panels is irrelevant without context. Both 655 and 656 are clearly brief teasers.

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Jackhartmann

Jackhartmann

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Posted
On 2/10/2023 at 2:39 PM, Jesse-Lee said:

I'm no scientist, but I'd guess it'd probably be 401 degrees Celsius.

To calculate the final temperature after thermal equilibrium is reached, we need to consider the heat exchange between the two pieces of bismuth. We can start by using the heat equation:

ΔQ = mcΔT

where ΔQ is the heat exchange, m is the mass, c is the heat capacity, and ΔT is the change in temperature.

We can first calculate the heat absorbed by the solid bismuth, which is given by:

ΔQ1 = mc1ΔT1 = (27.7 g)(26.3 J K–1 mol–1)(253 °C - Tf)

where Tf is the final temperature.

Next, we can calculate the heat released by the liquid bismuth, which is given by:

ΔQ2 = -mc2ΔT2 = -(277 g)(31.6 J K–1 mol–1)(Tf - 333 °C)

Since the two pieces of bismuth are in thermal equilibrium, the heat absorbed by the solid must be equal to the heat released by the liquid, so we can set ΔQ1 = ΔQ2:

(27.7 g)(26.3 J K–1 mol–1)(253 °C - Tf) = -(277 g)(31.6 J K–1 mol–1)(Tf - 333 °C)

Solving for Tf, we find:

Tf = (253 °C - 333 °C)(27.7 g)(26.3 J K–1 mol–1) / [(277 g)(31.6 J K–1 mol–1)] + 333 °C

Tf = 324.4 °C

So the final temperature after thermal equilibrium is reached is 324.4 °C.

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Brock

Brock

Posted
Posted
On 2/12/2023 at 8:39 PM, Jackhartmann said:

To calculate the final temperature after thermal equilibrium is reached, we need to consider the heat exchange between the two pieces of bismuth. We can start by using the heat equation:

ΔQ = mcΔT

where ΔQ is the heat exchange, m is the mass, c is the heat capacity, and ΔT is the change in temperature.

We can first calculate the heat absorbed by the solid bismuth, which is given by:

ΔQ1 = mc1ΔT1 = (27.7 g)(26.3 J K–1 mol–1)(253 °C - Tf)

where Tf is the final temperature.

Next, we can calculate the heat released by the liquid bismuth, which is given by:

ΔQ2 = -mc2ΔT2 = -(277 g)(31.6 J K–1 mol–1)(Tf - 333 °C)

Since the two pieces of bismuth are in thermal equilibrium, the heat absorbed by the solid must be equal to the heat released by the liquid, so we can set ΔQ1 = ΔQ2:

(27.7 g)(26.3 J K–1 mol–1)(253 °C - Tf) = -(277 g)(31.6 J K–1 mol–1)(Tf - 333 °C)

Solving for Tf, we find:

Tf = (253 °C - 333 °C)(27.7 g)(26.3 J K–1 mol–1) / [(277 g)(31.6 J K–1 mol–1)] + 333 °C

Tf = 324.4 °C

So the final temperature after thermal equilibrium is reached is 324.4 °C.

So if I’d said tree fiddy, I’d have been close?

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