A 60.5g piece of lead is at 18.6oC. If 328 cal of heat is added to the lead, what is its final temperature? Do not enter units.
I was trying to use 328 as the Q value and solve for delta T
HOWEVER, this wasn't working.
The Attempt at a Solution
Could someone please explain to me what equation I need to use and how the 328 cal are used? Thanks.