You add 47.0 grams of a metal to 96.0 grams of water. The metal began at a tempe
ID: 548497 • Letter: Y
Question
You add 47.0 grams of a metal to 96.0 grams of water. The metal began at a temperature of 84.0°C, and the water began at a temperature of 25.0°C. If the metal has a specific heat of 0.897 J/g°C and water has a specific heat of 4.18 J/g°C, what is the final temperature of the water in degrees celsius?
Hints: the energy lost by the metal is equal and opposite to the energy gained by the water, and the final temperature of both metal and water will be the same.
You add 47.0 grams of a metal to 96.0 grams of water. The metal began at a temperature of 84.0°C, and the water began at a temperature of 25.0°C. If the metal has a specific heat of 0.897 J/g C and water has a specific heat of 4.18 J/goC what is the final temperature of the water in degrees celsius? Hints: the energy lost by the metal is equal and opposite to the energy gained by the water, and the final temperature of both metal and water will be the same SUBMIT ANSWER search laExplanation / Answer
heat lost by metal = heat gained by water
mmetal*smetal*DT = mwater*swater*DT
47*0.897*(84-x) = 96*4.184*(x-25)
x = final temperature of mixer = 30.6 C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.