Question

What is the specific heat capacity of a 50-gram piece of 100°C metal that will change 400 grams of 20°C water to 22°C?

Answer 1

`s_("metal")=0.86J/(g*""^@C)`

Explanation:

According to the law of conservation of energy that sates:
Energy cannot be created nor destroyed, we can conclude that the energy lost from the metal is absorbed by water, and therefore;

`q_(lost)=q_("gained")`

in here, the metal will lose heat and water will gain heat. Note that the specific heat capacity of water is `s=4.18J/(g*""^@C)`.

Thus, `mxxs_("metal")xxDeltaT = mxxs_(water)xxDeltaT`

`50cancel(g)xxs_("metal")xx(100cancel(""^@C)-22cancel(""^@C))=400cancel(g)xx4.18J/(g*""^@C)xx(22cancel(""^@C)-20cancel(""^@C))`

Now we can solve for the specific heat capacity of the metal and we get:

`=>s_("metal")=0.86J/(g*""^@C)`

Here is a video that explains this topic with more details:
Thermochemistry | Enthalpy and Calorimetry.