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forsale [732]
3 years ago
11

2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, cal

culate the specific heat of aluminum.
Chemistry
1 answer:
NeTakaya3 years ago
8 0
Q =  M * C *ΔT

Q / <span>ΔT  = M

</span>Δf - Δi =  98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2

C = 1137 / 5068

C = 0.224 J/gºC</span>
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Read 2 more answers
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
Explain the term "amphoteric"
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n(CuSO₄) = 0.5 mol; amount of substance.
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M - molar mass.
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