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:

.

the answer is a. co2, this would be colder than the : )

explanation:

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current