A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

calorimeter and the water have a combined mass of 250.0 g and an overall specific heat of 1.035 cal/g•°C. The initial temperature of the calorimeter is 10.00°C. The system reaches a final temperature of 11.08°C when the metal is added.

Chemistry

1 answer:

V125BC [204] · Answer 1 · 2023-08-12T05:41:50+03:00

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

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.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 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

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$