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A student dissolves 4.51 grams of sodium hydroxide in 100.0 mL of water at 19.5 °C in a calorimeter. As the sodium hydroxide dissolves, the temperature of the surrounding water increases to 31.7 °C. What is the heat involved in the dissolving of the sodium hydroxide (the q)? Omg help meeeeee I’m in honors chem!

Sagot :

The heat involved in the dissolution of sodium hydroxide in the 100 mL of water is 5104.48 J

Data obtained from the question

  • Mass of NaOH = 4.51 g
  • Volume of water = 100 mL
  • Initial temperature of water (T₁) = 19.5 °C
  • Final temperature of water (T₂) = 31.7 °C
  • Heat of dissolving NaOH =?

How to determine the heat

Heat lost by NaOH = heat gained by the water

Thus, we shall determine the heat gained by the water in order to obtain the heat of dissolution of NaOH.

The heat gained by the water can be obtained as follow:

  • Mass of water (M) = 100 g
  • Initial temperature of water (T₁) = 19.5 °C
  • Final temperature of water (T₂) = 31.7 °C
  • Change in temperature (ΔT) = 31.7 – 19.5 = 12.2 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Heat (Q) =?

Q = MCΔT

Q = 100 × 4.184 × 12.2

Q = 5104.48 J

Thus, the heat of dissolution of NaOH is 5104.48 J

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