What is the total number of Joules of heat absorbed by 65.0 grams of water when the temperature of the water is raised from 25°C to 40°C?

Respuesta :

Answer:

Total number of heat absorbed is 4.08kJ

Explanation:

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The total amount of heat absorbed by the water is 4079.4 J

We'll begin by calculating the the change in the temperature of the water.

  • Initial temperature of water (T₁) = 25 °C
  • Final temperature (T₂) = 40 °C
  • Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 40 – 25

ΔT = 15 °C

Finally, we shall determine the heat absorbed by the water.

  • Mass of water (M) = 65 g
  • Change in temperature (ΔT) = 15 °C
  • Specific heat capacity of water (C) = 4.184 J/gºC
  • Heat absorbed (Q) =?

Q = MCΔT

Q = 65 × 4.184 × 15

Q = 4079.4 J

Therefore, the amount of heat absorbed by the water is 4079.4 J

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