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Metal: Specific Heat (J/goC)

Calcium: 0.647

Iron: 0.449

Silver: 0.235

Gold: 0.129

After absorbing 0.500 kJ of energy, a 55.5 g sample of an unknown substance has a temperature increase of about 14 degrees Celsius. Identify the substance.

Group of answer choices:

A.) Gold

B.) Iron

C.) Silver

D.) The substance is not listed.

E.) Calcium

Respuesta :

Identified substance which is present in the sample of 55.5g is calcium, because the calculated and actual specific heat is almost same.

What is the relation between specific heat and absorbed heat?

Relation between the amount of absorbed heat and specific heat of any substance is shown by following equation:

Q = mcΔT, where

  • Q = absorbed heat = 0.500kJ = 500J
  • m = mass of unknown substance = 55.5 g
  • c = specific heat
  • ΔT = change in temperature = 14 degrees Celsius

On putting all these values on the above equation, we get

c = 500 / (55.5)(14) = 0.643 J/g. degrees Celsius

And this value is almost equal to the given specific heat value of calcium, so the given substance is calcium.

Hence the given unknown substance is calcium.

To know more about specific heat, visit the below link:

https://brainly.com/question/1401668

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