Asked by Nevaeh

A 30.5 g sample of an alloy at 94.0°C is placed into 48.7 g water at 20.3°C in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g×K)

Answers

Answered by DrBob222
[mass alloy x specific heat alloy x (Tfinal-Tinitial)] + [(calorimeter constant x (Tfinal-Tinitial)] = 0
Answered by Nevaeh
(30.5)(X)(10.8)+(?)(10.8)=0

What number goes in the calorimeter constant ?
Answered by DrBob222
9.2 J/K = heat capacity of the coffee cup.
Answered by Nevaeh
It's still not giving me the right answer. I'm pulling in the numbers, and doing the equation right. But, it keeps telling me, I am wrong.
Answered by DrBob222
I didn't include the water and should have done so.
(mass water x specific heat water x (Tfinal-Tinitial)) + (mass alloy x specific heat alloy x (Tfinal-Tinitial)) + Ccal(Tfinal-Tinitial) = 0
Answered by uchenna
so, what is the t final- t initial for ccal?
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