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A sample of water with a mass of 123.00 kg is heated from 25 C to 97 C. If the specific heat of water is 1 J-1 kg K-1 then how much energy is required?

Respuesta :

Edufirst

Answer:

  • 8,860 J = 8.86 kJ

Explanation:

The heat energy required to heat a substance may be calculated from the formula:

  • Q = m × C × ΔT

Where:

  • m = mass of substance = 123.00 kg of water
  • C = specific heat = 1 J⁻¹ kg K⁻¹
  • ΔT increase of temperature = 97°C - 25°C = 72°C = 72 K.

Then, substitute the numbers in the formula to get:

  • Q = 123.00 kg × 1 J⁻¹ kg K⁻¹ × 72 K = 8,856 J ≈ 8,860 J or 8.86 kJ

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