Answer: 8.2689J/gC
H=mc(change in temp)
H(water)=50*4.184*3.32=694.5J
H(zinc)=694.5J
c(zinc)=694.5/(25.3*3.32)=8.2689J/gC
Taking into account the expressions and the principle of calorimetry that ensures that the body with a higher temperature will give up thermal energy to the one with a lower temperature until both are in an identical energy state, the specific heat of the zinc is 0.383 [tex]\frac{J}{gC}[/tex].
In first place, you need to know that the Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The equation for calculating heat exchanges results from the product of the specific heat times the mass of the body and the temperature difference. That is, the amount of heat that a body receives or transmits is determined by:
Q = c×m×ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (being ΔT= Tfinal - Tinitial)
On the other hand, a calorimeter is a device used to measure the heat of flow of a chemical reaction or physical change. It consists of a container built with insulating walls that limit heat exchanges with the environment in which the substances that give and receive heat are found. Being in a container with insulating walls that do not allow the exchange of energy with the outside, a principle ensures that the body with a higher temperature will give up thermal energy to the one with a lower temperature until both are in an identical energy state. That is, the heat change in the system must be 0. This is:
Qsystem= Qabsorbed + Qyielded
0= Qabsorbed + Qyielded
In this case, the water, when going from a higher temperature to a lower one, delivers a quantity of heat. Then
Qyielded= Qwater= cwater×mwater ×ΔT= 4.184[tex]\frac{J}{gC}[/tex] × 50 g× (96.68 C - 100 C)
On the other hand, the heat given off by the water will be absorbed by the zinc. Then:
Qabsorbed= Qzinc= czinc×mzinc×ΔT= czinc× 25.3 g× (96.68 C - 25 C)
Taking into account the principle mentioned above, it is possible to express:
0= czinc× 25.3 g× (96.68 C - 25 C) + 4.184[tex]\frac{J}{gC}[/tex] × 50 g× (96.68 C - 100 C)
Solving:
0= czinc× 25.3 g× (96.68 C - 25 C) + 4.184[tex]\frac{J}{gC}[/tex] × 50 g× (96.68 C - 100 C)
0= czinc× 25.3 g× 71.68 C + 4.184[tex]\frac{J}{gC}[/tex] × 50 g× (-3.32 C)
0= czinc× 1813.504 g×C - 694.544 J
694.544 J= czinc× 1813.504 g×C
[tex]\frac{694.544 J}{1813.504 gC} =czinc[/tex]
0.383 [tex]\frac{J}{gC}[/tex] =czinc
In summary, the specific heat of the zinc is 0.383 [tex]\frac{J}{gC}[/tex]
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