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When 3000 g of water is cooled from 80.0C to 10.0C, how much heat energy is lost?

Sagot :

boffeemadrid

Answer:

[tex]878640\ \text{J}[/tex]

Explanation:

m = Mass of water = 3000 g

c = Specific heat of water = [tex]4184\ \text{J/kg}^{\circ}\text{C}[/tex]

[tex]\Delta T[/tex] = Change in temperature of water = [tex](10-80)^{\circ}\text{C}[/tex]

Heat is given by

[tex]Q=mc\Delta T\\\Rightarrow Q=3\times 4184\times (10-80)\\\Rightarrow Q=-878640\ \text{J}[/tex]

Heat energy lost is [tex]878640\ \text{J}[/tex].

Lanuel

The quantity of heat lost is equal to 877,800 Joules or 8778 Kilojoules.

Given the following data:

  • Mass of water = 3000 grams.
  • Initial temperature = 10.0°C.
  • Final temperature = 80.0°C.

Scientific data:

  • Specific heat capacity = 4.18 J/g °C.

How to calculate the quantity of heat lost.

Mathematically, the quantity of heat lost 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 = 3000 \times 4.18 \times (80-10)\\\\Q = 3000 \times 4.18 \times70[/tex]

Q = 877,800 Joules or 8778 Kilojoules.

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

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