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OleMash [197]
3 years ago
13

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 ha

s been reached, the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from?
Chemistry
1 answer:
Tju [1.3M]3 years ago
6 0
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
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This question can be simply solved by using heat formula,
    Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
 Q = 456 g x 4.186 J / g K x 25 K
     = 47720.4 J
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hence 47.72 kJ of heat energy should be removed.
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3 years ago
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Explanation:

Hello!

In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

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