DAT Destroyer GChem Question #201

[emminent]

Advertisement - Members don't see this ad

So I searched and found someone posted a question but it didn't answer my question about it.

Question 201 Gchem:
An Unknown metal of 50 grams is initially at 100C and dropped into a beaker of 200 grams of water at 30C. The final temperature of the system is 40 C. Find the specific heat of the metal if the heat of water is 4.2 J/g C.

Answer is setup is mCdeltaT(water) = mCdeltaT (water)

Romano put down:
50*c*(100-40) = 200*4.2*(40-30)
Everything seems right to me except temperature change for 100-40.

Isn't the initial temperature 100 and the final 40 for the metal? Thus shouldn't it be 40-100? But then again specific heat of metal would be negative... wth?

 

[discensdentibus]

you are missing a key detail:
He sets it up as "Heat LOST by metal" = "Heat GAINED by water"

(-) qmetal= qwater b/c qmetal+qwater = 0
but he can keep both delta Ts positive by flipping the sign on the heat of the metal. Really, heat lost is represented as a negative value, but he assumes you would know that and simplifies the calculation.

 

[emminent]

Ah... much thanks

 

[discensdentibus]

NP its all a part of the learning process