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A calorimeter contains 100 g of water at 39.8 ºC. A 8.23 g object at 50 ºC is placed inside the calorimeter. When equilibrium has been reached, the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from?

Respuesta :

superman1987
when equilibrium has been reached so, according to this formula we can get the specific heat of the unknown metal and from it, we can define the metal as each metal has its specific heat:

Mw*Cw*ΔTw = Mm*Cm*ΔTm

when 
Mw → mass of water
Cw → specific heat of water
ΔTw → difference in temperature for water 

Mm→ mass of metal
Cw→ specific heat of the metal
ΔTm → difference in temperature for metal

by substitution:

100g * 4.18 * (40-39.8) = 8.23 g * Cm * (50-40)

∴ Cm = 83.6 / 82.3 = 1.02 J/g.°C

when the Cm of the Magnesium ∴ the unknown metal is Mg

The object in the calorimeter which is burnt is Magnesium.

In the reaction, it is assumed that no energy is lost or gained.

So, [tex]\rm \Delta E[/tex] water = [tex]\rm \Delta E[/tex] metal

[tex]\rm m_1c_1\Delta T[/tex] = [tex]\rm m_2c_2\Delta T[/tex]

100 [tex]\times[/tex] 4.814 [tex]\times[/tex] (40 - 39.8) = 8.23 [tex]\times[/tex] specific heat of metal [tex]\times[/tex] (40 - 50)

specific heat of metal = -1.1692 J/[tex]\rm g^\circ C[/tex].

The calculated value is for the specific heat of Magnesium.

Thus the object in the calorimeter which is burnt is Magnesium.

For more information about specific heat, refer to the link:

https://brainly.com/question/11297584?referrer=searchResults

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