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°)​

 

[asteriatic]

I would approach this problem by first understanding the concept of specific heat and how it relates to the amount of energy required to raise the temperature of a substance. I would also take into consideration the heat capacity of the copper calorimeter and its impact on the overall energy of the system.

Using the given data and the equation Q=mcΔT, I can calculate the initial energy of the system before the addition of 100g of water. This would be equal to the energy gained by the water and the calorimeter, which can be represented as (65g+30g) x 386 J/kg-K x (49°C-20°C). This gives us an initial energy of 21,630 J.

Next, I would consider the energy added by the 100g of water at 80°C. Using the same equation, we can calculate the energy gained by this water as 100g x 4186 J/kg-K x (49°C-80°C), which is equal to -12,534 J. This negative value indicates that the water is losing energy as it cools down to the final temperature of 49°C.

Now, to find the specific heat of material X (Cx), we can set up an equation where the total energy gained by the system (initial energy + energy added by the 100g of water) is equal to the sum of the masses (75g+65g+30g) multiplied by the specific heat of water (4186 J/kg-K) and the specific heat of the copper calorimeter (386 J/kg-K) and the specific heat of material X (Cx) multiplied by the change in temperature (49°C-20°C). This can be represented as:

21,630 J - 12,534 J = (75g+65g+30g) x (4186 J/kg-K + 386 J/kg-K + Cx) x (49°C-20°C)

Solving for Cx, we get Cx = 4.2 J/kg-K. This is the specific heat of material X.

In conclusion, the specific heat of material X (Cx) is 4.2 J/kg-K. It is important to note that this value may vary depending on the purity and physical properties of the material. Further experimentation and analysis may be needed to accurately determine the specific heat of material X.