A 23.0 g piece of metal at 99.0 ∘C is placed in a calorimeter containing 53.2 g of water at 24.0 ∘C. The final temperature of the mixture is 26.1 ∘C. What is the specific heat capacity of the metal? Assume no energy is lost to the surroundings.

The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. Its units are J/g.ºC.

If we know the specific heat and the amount of a substance, then the change in the sample’s temperature (ΔT) will tell us the amount of heat (q) that has been absorbed or released in a particular process. The equation for calculating the heat change is given by:

[tex]: q=m.s.ΔT[/tex]

Where ΔT is the temperature change: [tex]ΔT= tfinal - tinitial[/tex], m the mass and s the specific heat.

If no energy is lost to the sorroundings, then all the heat lost by the metal will be absorbed by the water. Therefore, the heat change of the system (qsystem) must be zero and we can write:

[tex]qsystem = qwater + qmetal[/tex]

[tex]qwater = -qmetal[/tex]

Replacing each term with the equation for calculating heat change:

[tex]mw.sw.ΔTw = -[mm.sm.ΔTm ][/tex]

A recommendation is to carry the units through the entire sequence of calculations. Therefore, if the equation is set up correctly, then all the units will cancel except the desired one.

[tex]53.2 g . 4.184 J/g°C . (26.1 - 24.0)ºC = -[23.0 g . sm . (26.1 - 99.0)°C][/tex]

[tex]464.436 J = -[23.0 g . sm . (-72.9)°C][/tex]

[tex]sm = 464.436 J/ -[-1676.7 g°C][/tex]

[tex]sm = 0.277 J/g.°C[/tex]

Thus, the specific heat of the metal is 0.277 J/g.ºC.