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?

1 Answer
May 20, 2016

Answer:

#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.