Solving for cx: Specific Heat of Material X | Calorimetry Question

[davekardle]

Homework Statement

To find cx, the specific heat of material X, I place 75g of X in a 30g copper
calorimeter that contains 65g of water, all initially at 20°C. When I add 100g of
water at 80°C, the final temperature is 49°C. What is cx?
Data: cCU = 386 J kg
-1
K
-

Homework Equations

Q=mcDt

The Attempt at a Solution

I tried finding out the amount of energy around in the system of water and cupper and then added the energy provided by the addition of X, with its variable Cx.
I then added the energy provided by the 100g of water at 80. and equalled that to the ( Sum of masses x Cw x Cc x Cx x 49. and then tried to calculate the variable Cx by rearranging. The answer is 2.18 Kj. I can't find it :(

 

[tiny-tim]

welcome to pf!

hi davekardle! welcome to pf!

show us what you've tried, and then we'll know how to help!

 

[davekardle]

Energy Available in the system ( H2O + Cu)

Q= 0.03 x 0.386 x 20 = 0.2316 KJ (Cu)
Q= 0.065 x 4.2 x 20 = 5.5

total Qw + Qcu = 5.7 Kj KG- C-

ENERGY offered by X:

Q= 0.075x Cx x 20= 1.5Cx

Energy offered by water:

Q= 0.1 x 4.2 x 80 = 33.6 Kj Kg- c-

ADDING energies=

5.5 + 1.5Cx + 33.6 = 0.170kg(TOTAL mass of mix) x (0.27 x 4.2) x Cx x 49 ( final temp)


rearranging

39.3 = 19.9Cx
Cx= 1.97Kj

Which is wrong :(

 

[tiny-tim]

i don't understand this bit …

0.170kg(TOTAL mass of mix) x (0.27 x 4.2) x Cx x 49 ( final temp)

(what is the 0.27 ? and …)

don't you need to continue to treat each of the materials separately?

(btw, you've used a "zero-energy-level" of 0° …

it would be easier and quicker to use 20°)​