Chemistry Class Question...?
I have a question on this question for my chem class. The iniitial question is 45.5 g of copper at 99.8 deg C is dropped into a beaker containing 152g of water at 18.5 C. What is the fianl temp when the thermal equilibrium is reached? Now i have it down to [(4.184 J/g K)(152g)(T Final-291.5K) + (.385J/g)(45.5g)(T final-372.8K)]=0, but I'm stumped when it comes to finding the final temp.. Please help.
Answer:
remember at equilibrium, the temperatures are the same so....
[(4.184 J/g K)(152g)(T Final-291.5K) + (.385J/g)(45.5g)(T final-372.8K)]=0
subtract (.385J/g)(45.5g)(T final-372.8K) from both sides
(4.184 J/g K)(152g)(T Final-291.5K)= -(.385J/g)(45.5g)(T final-372.8K)
multiplying it out gives us (i rounded to tenths)
635.9*Tf - 185385 = -17.5Tf + 6530.5
getting Tf on same side...
653.5*Tf = 191915.2
and
Tf = 293.7
Hope thats right!Can someone please tell me if theres a scientific name for glass like h20 is for water?
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Answer:
remember at equilibrium, the temperatures are the same so....
[(4.184 J/g K)(152g)(T Final-291.5K) + (.385J/g)(45.5g)(T final-372.8K)]=0
subtract (.385J/g)(45.5g)(T final-372.8K) from both sides
(4.184 J/g K)(152g)(T Final-291.5K)= -(.385J/g)(45.5g)(T final-372.8K)
multiplying it out gives us (i rounded to tenths)
635.9*Tf - 185385 = -17.5Tf + 6530.5
getting Tf on same side...
653.5*Tf = 191915.2
and
Tf = 293.7
Hope thats right!