What is the Cp (in J/gC) of a material if a 50 gram sample of it at 100 C is placed into 500 grams of water at 20 C and the water temp rises to 23 C and the material temp goes down to 23 C?

The specific heat capacity of this material is approximately .

(Assume that there is no energy loss. Assume that the specific heat capacity of water is .)

Explanation:

Let denote the specific heat capacity of a material. Consider a sample of that material with mass . If the temperature of that objects changes by , the corresponding energy change would be .

Start by finding the amount of energy that water has gained from the sample.

That of water would have gained:

.

Assume that the sample and the of water were insulated from the surroundings. The energy that water gained would be exactly the same as the energy that the sample had released through cooling.

In other words, the sample has released approximately of energy as it is cooled from to .

Rearrange the equation to find an expression for specific heat capacity, :

.

Apply this expression to find the specific heat capacity of the sample:

.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles