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A student wants to calculate the heat gained of 100 grams of water in order to
calculate the specific heat of a metal in the following problem. The water is
heated from 25C to 50C when the metal is added. The specific heat of water is
4.18 J/gºC. Use Q=m xCxT*

Respuesta :

Lanuel

The quantity of heat gained to raise the temperature from 25.0°C to 50.0°C is 10,450 Joules.

Given the following data:

  • Mass of water = 100 grams
  • Initial temperature = 25.0°C
  • Final temperature of water = 50.0°C
  • Specific heat capacity of water = 4.18 J/g°C

To find the quantity of heat gained:

Mathematically, heat capacity or quantity of heat is given by the formula;

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

Where:

  • Q represents the heat capacity or quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

Substituting the values into the formula, we have:

[tex]Q = 100(4.18)(50 - 25)\\\\Q = 418(25)[/tex]

Quantity of heat = 10,450 Joules.

Read more: https://brainly.com/question/188778

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