Asked by Rosie

A 1.50 X 10^2g piece of brass(specific heat capacity 3.80 X 10^2 J(kgC) is submerged in 400 mL of water at 27.7 Degrees Celsius. What is the original temperature of the brass if the mixture has a temperature of 28.0 degrees Celsius
Answered by bobpursley
haat added to brass+heat added to water=0
150*cbrass*(28-Ti)+400*cwater*(28-27.7)=0
solve for Ti
Answered by Michelle
would the answer by 3827
Answered by Ms. Sue
Michelle/Rosie -- please use the same name for your posts.
Answered by Rosie
i think the answer would be 3827.1
Answered by bobpursley
absolutely several thousand off.
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