What mass of water can be heated from 25.0° C to 50.0° C by the addition of 2825 J of heat energy if the specific heat capacity of water is 4.184 J/goC?
35 g
13 g
27 g
20 g

Respuesta :

Lanuel

Answer:

Mass, m = 27g

Explanation:

Given the following data;

Initial Temperature, T1 = 25°C

Final temperature, T2 = 50°C

Quantity of heat = 2825J

Specific heat capacity of water = 4.184

Heat capacity is given by the formula;

[tex] Q = mcdt[/tex]

Where,

  • Q represents the heat capacity.
  • m represents the mass of an object.
  • c represents the specific heat capacity of water.
  • dt represents the change in temperature.

Making mass, m the subject of formula, we have;

[tex] m = \frac {Q}{cdt} [/tex]

Change in temperature, dt = T2 - T1

Change in temperature, dt = 50-25 = 25°C

Substituting into the equation, we have;

[tex] m = \frac {2825}{4.184 * 25} [/tex]

[tex] m = \frac {2825}{104.6} [/tex]

Mass, m = 27g

Therefore, the mass of water that can be added is 27 grams.

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