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The quantity of heat that is required to raise this sample of liquid water is 20,900 Joules.

Given the following data:

  • Mass of liquid water = 50.0 grams.
  • Initial temperature = 50°C
  • Final temperature = -50°C

Scientific data:

  • Specific heat capacity of water = 4.18 J/g °C.
  • Specific heat capacity of ice = 2.108 J/g °C.

How to calculate the quantity of heat.

Mathematically, quantity of heat is given by this formula:

[tex]Q = mc \theta[/tex]

Where:

  • m is the mass.
  • c is the specific heat capacity.
  • [tex]\theta[/tex] is the change in temperature.

Substituting the given parameters into the formula, we have;

[tex]Q=50.0 \times 4.18 \times (-50-50)\\\\Q=209(-100)[/tex]

Quantity of heat = -20,900 Joules.

Read more on heat capacity here: brainly.com/question/16559442

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