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A 200 g block of a substance requires 520 J of heat to raise its temperature from 25°C to 45°C.

Use the table to identify the substance.

Respuesta :

Answer:

The answer is gold.

Explanation:

Substance: Specific Heat (C) in joules per gram/degrees centigrade:

Water (ice) 2.05

Iron 0.46

Aluminium 0.90

Gold 0.13

Copper 0.39

Ammonia (liquid) 4.70

Ethanol 2.44

Gasoline 2.22

Water (liquid) 4.18

Water (vapor) 2.08

Air (25 degrees Celsius) 1.01

Oxygen 0.92

Hydrogen 14.30

To know that we clear the equation q=m*C*Δt for the specific heat (C) and solve the equation. Keep in mind that we need to use the heat (520 J) in Joules since the values from the question and the table are in Joules.

q/m∗ΔT = C = 520J/200g∗(45degC−25degC) =

0.13J/gdegC

As you can see, the substance that has this specific heat is GOLD.

The value of specific heat for the given data in the question will be for gold:-

[tex]C= 0.13\ \dfrac{J}{g^oC}[/tex]

What will be the Specific heat of the material?

Specific heat can be defined as the amount of heat one kg or gram substance can absorb or release for a unit temperature difference.

The formula for the heat will be:-

[tex]Q=mc\Delta T[/tex]

m= Mass of substance

c= specific heat of the substance

[tex]\Delta T[/tex]= Temperature difference

Now the given data is

m= 200 g

Q= 520 J

[tex]\Delta T[/tex]=  (45-25)=20

[tex]Q=mc\Delta T[/tex]

[tex]520=200\times C\times 20[/tex]

[tex]C= \dfrac{520}{200\times 20} =\dfrac{520}{4000} =0.13\ \dfrac{J}{g ^oC}[/tex]

Since the value of specific gold  is  [tex]0.13\ \dfrac{J}{g ^oC}[/tex]

Thus the value of specific heat for the given data in the question will be for gold:-

[tex]C= 0.13\ \dfrac{J}{g^oC}[/tex]

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