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If the temperature of 96.5 g of water increases from 25.2 YC to 37.6 YC, how much heat did the water absorb?

Respuesta :

sebassandin

Answer:

Q = 5,006.6 J

Explanation:

Hello there!

In this case, for these calorimetry problems, it is possible for us to compute the heat absorbed or released by a substance by considering the involved mass, specific heat and temperature change:

[tex]Q=mC(T_2-T_1)[/tex]

Thus, since the mass of water is 96.5 g, its specific heat is widely known as 4.184 J/g°C and the temperature increases, the resulting heat turns out to be:

[tex]Q=96.5g*4.184\frac{J}{g\°C} (37.6\°C-25.2\°C)\\\\Q=5,006.6J[/tex]

Best regards!

Cricetus

The heat absorbed by the water will be "5006.6 J".

The given values are:

Temperature,

  • [tex]T_1 = 25^{\circ} C[/tex]
  • [tex]T_2 = 37.6^{\circ} C[/tex]

then,

  • [tex]\Delta T = T_2-T_1[/tex]

              [tex]= 37.6-25.2[/tex]

              [tex]= 12.4^{\circ} C[/tex]

Mass,

  • [tex]m = 96.5 \ g[/tex]

As we know the formula,

→ [tex]Q = mC\Delta T[/tex]

By substituting the values, we get

      [tex]= 96.5\times 4.184\times 12.4[/tex]

      [tex]= 5006.6 \ J[/tex] (required heat)

Thus the response above is right.

Learn more about heat here:

https://brainly.com/question/15334876

Ver imagen Cricetus

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