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The temperature of a 24.5-g block of aluminum decreases from 30.0°C to 21.5°C. If aluminum has a specific heat of 897 J/(kg-°C), how large is the change in thermal energy of the block of aluminum?

A. 187 J

C. 5,820 J

B. 2,590 J

D. 187,000 J

Respuesta :

Eduard22sly

The change in thermal energy of the block of aluminum, given it decreases from 30.0°C to 21.5°C is 187 J (Option A)

How to determine the change in thermal energy

We'll begin by obtaining the temperature change of the aluminum. This is obtained as follow:

  • Initial temperature (T₁) = 30.0 °C
  • Final temperature (T₂) = 21.5 °C
  • Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 21.5 – 30

ΔT = –8.5°C

Finally, we shall determine the change in thermal energy. This can be obtained as follow:

  • Change in temperature (ΔT) = –8.5°C
  • Mass of aluminum (M) = 24.5 g = 24.5 / 1000 = 0.0245 Kg
  • Specific heat capacity of aluminum (C) = 897 J/KgºC
  • Change in thermal energy (-Q) =?

-Q = MCΔT

-Q = 0.0245 × 897 × -8.5

-Q = -187 J

Q = 187 J

Thus, the change in thermal energy is 187 J (Option A)

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