Question
A 95 gram piece of metal is heated to 98°C and placed into 200mL of water insulated at 20°C. If the equilibrium temperature is 27.20, find:
The heat gained by the water. calories
The specific heat of the metal. cal/(g°C)
Hint: 1 mL water has a mass of 1 gram
I have no idea how to begin any of this, much less how to determine the answers for the two parts. Can someone help please?
The heat gained by the water. calories
The specific heat of the metal. cal/(g°C)
Hint: 1 mL water has a mass of 1 gram
I have no idea how to begin any of this, much less how to determine the answers for the two parts. Can someone help please?
Answers
bobpursley
heat gained by water:
masswater*specificheat*deltatemp
200g*4.18j/g*(27.2-20)=...
heat gained by water+heat change by metal=0
above heat calculation+95g*cmetal*(27.2-98)=0
solve for cmetal
masswater*specificheat*deltatemp
200g*4.18j/g*(27.2-20)=...
heat gained by water+heat change by metal=0
above heat calculation+95g*cmetal*(27.2-98)=0
solve for cmetal
Anna
6,019.2 calories for the heat gained by water is incorrect. That has been my problem all along. Can't figure out the correct answer.
Thanks, though.
Thanks, though.