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

Answer 1

haat added to brass+heat added to water=0
150*cbrass*(28-Ti)+400*cwater*(28-27.7)=0
solve for Ti

Answer 2

would the answer by 3827

Answer 3

Michelle/Rosie -- please use the same name for your posts.

Answer 4

i think the answer would be 3827.1

Answer 5

absolutely several thousand off.