50g*4.184J/g*C*(Tf-100)+50g*4.184J/g*C*(Tf-25)=0
50g*4.184J/g*C*Tf-4184J-50g*4.184J/g*C*Tf+1046J=0
-3138J=0
Tf=3138J/50g*4.184J/g*C
Tf=37.7°C
how would i calculate; a 50g sample of water at 100 degrees is poured into a 50g sample of water at 25 degrees. with a specific heat of water (l) being 4.184 j/g degrees C.what will be the final temperature of the water?
heat gained + heat lost = 0
[massH2O*specificheat*(Tf-Ti)+[massH2O*specific heat*(Tf-Ti)].
Tf is final T. Solve for this.
Ti is initial T of each.
Post your work if you get stuck and need further assistance. There is a short cut way of doing this but the above equations works for almost anything.
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