A 10 g sample of an unknown metal is cooled from 100 °C to 21.6 °C in a calorimeter of 100 g of water. The temperature of the wa

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A 10 g sample of an unknown metal is cooled from 100 °C to 21.6 °C in a calorimeter of 100 g of water. The temperature of the wa

ter rises from 20.0 °C to 21.6 °C. The specific heat of water is 4.184J/g°C. Calculate the specific heat of the unknown metal to determine the identity of the metal using the table below.
Metal Specific Heat (J/g°C) Magnesium 1.047

Aluminum 0.900

Tin 0.226

Nickel 0.443

Show work here: Specific heat of unknown metal: ___________ The metal is (circle one): Magnesium Aluminum Tin Nickel

Chemistry

1 answer:

Marta_Voda [28] · Answer 1 · 2022-11-19T16:53:21+03:00

Taking into account the definition of calorimetry, the specific heat of the unknown metal is 0.9 $\frac{J}{gC}$ and the metal is aluminum.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

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.

In this case, you know:

For unknown metal:

Mass of metal = 10 g
Initial temperature of metal= 100 °C
Final temperature of metal= 21.6 ºC
Specific heat of metal= unknown

For water:

Mass of water = 100 g
Initial temperature of water= 20 ºC
Final temperature of water= 21.6 ºC
Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

For unknown metal: Qmetal= $c_{unknown metal}$ × 10 g× (21.6 C - 100 C)

For water: Qwater= 4.186 $\frac{J}{gC}$ × 100 g× (21.6 C - 20 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- $c_{unknown metal}$ × 10 g× (21.6 C - 100 C)=4.186 $\frac{J}{gC}$ × 100 g× (21.6 C - 20 C)

Solving:

- $c_{unknown metal}$ × 10 g× (- 78.4 C)= 669.76 J

$c_{unknown metal}$ × 784 gC= 669.76 J

$c_{unknown metal}$ = 669.76 J÷ 784 gC

$c_{unknown metal}$ = 0.854 $\frac{J}{gC}$ ≅ 0.9 $\frac{J}{gC}$

Finally, the specific heat of the unknown metal is 0.9 $\frac{J}{gC}$ and the metal is aluminum.

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