(DrBob222 please help me)
What would be the final temperature of the system if 16.6 g of lead at 138 ◦C is dropped into 23.4 g of water at 8.68 ◦C in an insulated container? The specific heat of lead is 0.128 J/g◦C.
Please explain with the numbers plugged in to the equation
2 answers
[16.6(0.128)(t-138]+[23.4(4.184)(t-8.68)]=0 Solve for t.
Couldn't you have followed the Pb/water problem Bob P worked for you?