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?

Question

Rosemary

Chemistry

9 January, 13:04

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?

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Answers (1)

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Gary Cobb · Answer 1

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