### #33

greenspun.com : LUSENET : UR General Chemistry : One Thread

How do you go about doing #33 in the book (chapter 9)? It's probably really easy but I'm lost....

-- Anonymous, February 06, 2000

This is probably a complicated version of doing this problem, but here goes......

We are attempting to calculate the specific heat of copper. We know that the units for specific heat are Joules/ C g.

First, work with what you are given....DeltaT= (94.4-21.8= 73.6) and mass of Cu= 46.2 g Plug these numbers into the formula Joules/ C g where C= temperature in degrees Celsius and g is mass in grams. Now all you need to find is Energy, since you know that Energy is measured in Joules( and this is the unit which you need in order to complete the formula for finding the s. heat. You can obtain energy from the formula deltaE= 3/2RdeltaT. --> (3/2)(8.31)*73.6)=E= 917.424J. Now plug this energy into the formula for s.h. and you get- ->

917.424J/(46.2g)(73.6)= spec. heat specific heat= .269 J/Celsius grams

Essentially, I solved this problems by using the units of s.h. as a guide. I think this should be correct...

-- Anonymous, February 06, 2000

Or another way;

delta = d; marking the values for copper with a "c" and the values for water with a "w". (ex. mass of water = mw). Specific heat of water is Cw and for copper Cc.

heat gain by water = heat lost by copper

mc*Cc*dTc = mw*Cw*dTw

Cc = (mw*Cw*dTw) / (mc*dTc)

-- Anonymous, February 06, 2000