A 70g piece of ice at 0 Celsius is added to a sample of water at 6 Celsius. All of the ice melts and the temp of the water decreases to 0 Celsius. How many grams of water were in the sample.

Question

anonymous · Answer

specific heat equation

mC(dt)=mC(dt)

dt = t_f-t_o
m is mass
C is specific heat

anonymous · Answer

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html Q must be equal to each other

anonymous · Answer

I'm still confused. Surprise, surprise!  This is like a foreign language to me.  I started out by multiplying 70g by 80.cal/1g.ice=5600 cal.  Is this even close?  I know it's not the only step.

anonymous · Answer

C does not mean call
C is the specific heat of the material

you need to look this up
the specific heat of ice and water are different

anonymous · Answer

specific heat of water
C_water= 4.186 joule/gram °C 
specific heat of ice 
 C_ice =2.108 kJ/kg-K

$$m_{ice}C_{ice}\Delta t=m_{water}C_{water}\Delta t$$

anonymous · Answer

ice goes from 0 to 6
so delta t = 6-0
water goes from 6 to 0
so delta t = 0-6

sorry forgot a negative sign in there

anonymous · Answer

Ok, so is this close?   70g*2.108kj*6=m*1.0*-6

anonymous · Answer

70*2.108*6=m*4.186*6

now solve for m

anonymous · Answer

35.25g??

anonymous · Answer

Well, that answer is wrong, so obviously I have NO idea what on earth i'm doing.

anonymous · Answer

Or you forgot the significant digit! It may 35,3g

anonymous · Answer

Nope, it's still wrong.  :(