Question

9.) 250.0g of copper at 100.0 degrees C are placed in a cup containing 325.0g of water at 20.0 degrees C. Assume no less to the surroundings. What is the final temperature of the copper and water?

So far I have used q=mc times delta t but I cannot find C which is the specific heat and therefore my formula isnt appearing right. Also after I do this, how would I implement the numbers so that I will get it for one gram, the temperature. I am a bit confused. THanks for your help.

Answer 1

loss of heat by copper + gain of heat by water = 0
massCu x specificheatCu x (Tf-Ti) + massH2O x specific heat water x (Tf-Ti) = 0
mass Cu = 250.0 g
sp. h. Cu = 0.385 J/g*C in my table.
Tf = final T Cu --solve for this
Ti Cu = 100 C.

mass H2O = 325.0 g
sp.h. H2O = 4.184 J/g*C
Tf = final T H2O --solve for this (note final T for Cu and final T for H2O is the same therefore there is only one unknown).
Ti H2O = 20.0 C.

You have only one unknown; i.e., Tf.