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Water (3010 g ) is heated until it just begins to boil. If the water absorbs 5.79x105 J of heat in the
process, what was the initial temperature of the water?

Respuesta :

dsdrajlin

Answer:

54 °C

Explanation:

Step 1: Given and required data

  • Mass of water (m): 3010 g
  • Heat absorbed (Q): 5.79 × 10⁵ J
  • Specific heat of water (c): 4.184 J/g.°C
  • Initial temperature (T₁): ?
  • Final temperature (T₂): 100 °C (Boiling point of water)

Step 2: Calculate the initial temperature of water

We will use the following expression.

Q = c × m × (T₂ - T₁)

5.79 × 10⁵ J = 4.184 J/g.°C × 3010 g × (100 °C - T₁)

T₁ = 54 °C