destroyer chem 185

[csulapredental]

An unknown metal of 50g is intially at 100C and dropped into a beaker of 200g of water at 30;C. The final temperature of the system is 40C. Find the specific heat of the metal if the specific heat of water is 4.2 J/g ;C.

Answer: 2.8 J/g ˚C


Heat lost by metal = Heat gained by H2O

m c deltaT = m c deltaT

(50)c(60) = (200)(4.2)(10) <--- having a problem here..

3000c = 8400

thus, c=2.8 J/gC


don't we always calculate the delta portion as final minus the initial? (40-100)

metal LOST heat which makes it -60, not +60?
heat gained by H2O seems to be +10, which is final-initial (40-30) giving a positive value of 10.

a negative value wasn't an option, so luckily i got this problem right.. but was just wondering how it works. i'm sure would've gotten it wrong if -2.8 was one of the options.

can someone help me out here? thanks.

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[Charles_Darwin]

Good thought - I don't think i've ever seen it that way though! I just assumed ΔT to mean the value of the difference: 60

 

[csulapredental]

Good thought - I don't think i've ever seen it that way though! I just assumed ΔT to mean the value of the difference: 60

its always final - initial tho aint it

 

[Lnguyen7]

Heat lost by metal = Heat gained by H2O

That's your problem there.
Assuming the system is isolated or no heat lost to the surrounding
qm + qH2O = 0
So qm = -qH2O

Now you can do DeltaT = final - initial (as always) and you will get a positive number

You were lucky indeed!