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?

Question

Noah Aldrich · Accepted Answer

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

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.

Noah Aldrich · Answer

undefined The specific heat capacity of the metal can be calculated using the formula:

Q = mcΔT

Where:
Q = heat transferred
m = mass
c = specific heat capacity
ΔT = change in temperature

First, calculate the heat transferred to the water:

Q_water = mcΔT = (400 g)(4.18 J/g°C)(22°C - 20°C) = 334.4 J

Since the heat transferred to the water is equal to the heat gained by the metal:

Q_metal = Q_water

Now, rearrange the formula to solve for the specific heat capacity of the metal:

c = Q_metal / (mΔT)

c = (334.4 J) / ((50 g)(100°C - 20°C))

c ≈ 0.67 J/g°C

So, the specific heat capacity of the metal is approximately 0.67 J/g°C.