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Chemistry 18 Online
OpenStudy (anonymous):

Which substance in Table 5.2 requires the smallest amount of energy to increase the temperature of 51.5g of that substance by 10K ? All of these are specific heats of some substances at 298 K. So N2(g) is 1.04 J/g-k Al (s) .90 J/g-k Fe (s) .45 J/g-k Hg (l) .14J/g-k H2O (l) 4.18 J/g-k CH4 (g) 2.20 J/g-k CO2 (g) .84 J/g-k CaCO3 (s) .82 J/g-k

OpenStudy (aaronq):

the one with the smallest specific heat capacity (which are the values you listed).

OpenStudy (anonymous):

Hg?

OpenStudy (aaronq):

according to your table, yes

OpenStudy (anonymous):

Okay so how do I calculate the energy needed for this temperature change.

OpenStudy (anonymous):

Cause this is all kinds of confusing to me.

OpenStudy (anonymous):

@aaronq

OpenStudy (aaronq):

Sorry, i'm not getting notifications. you can do it in several ways, but with this information you can use the calorimetry equation: \(q=m*C_P*\Delta T\) q= heat (energy) Cp= specific heat capacity m=mass dT=change in temp

OpenStudy (anonymous):

okay sooo I'm going to walk through this and see if I get it right.

OpenStudy (aaronq):

okay, you can post it here, i'll check it out

OpenStudy (anonymous):

dT would be 298-10K right since the table is in 298K and we're figuring out for 10K?

OpenStudy (anonymous):

q= .14 J/g-k * 51.5g * 288K q= 2076.48

OpenStudy (aaronq):

the change in temperature is only 10 K

OpenStudy (anonymous):

oh okay

OpenStudy (aaronq):

"increase the temperature of 51.5g of that substance by 10K"

OpenStudy (anonymous):

so only 72!

OpenStudy (anonymous):

72 joules

OpenStudy (aaronq):

yep!

OpenStudy (anonymous):

Thank you!

OpenStudy (aaronq):

no problem!

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