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What must be the specific heat of a sample of an unknown material of 36.359g,when 59.912J of heat are applied raising the temperature 152°C?

Respuesta :

nkirotedoreen46

Answer:

0.0108 J/g°C

Explanation:

We are given;

  • Mass of the unknown material as 36.359 g
  • Quantity of heat is 59.912 J
  • Temperature change, ΔT = 152°C

We are required to calculate the specific heat capacity of the material.

  • Note that the quantity of heat absorbed is calculated by multiplying mass by specific heat and by change in temperature.
  • Q = mcΔT

To calculate the specific heat capacity we rearrange the formula;

c = Q ÷ mΔT

  = 59.912 J ÷ (36.359 g × 152°C)

  = 0.0108 J/g°C

Therefore, the specific heat capacity of the unknown material is 0.0108 J/g°C

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