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If 1.13 x 104 J of heat is added to a water sample and the temperature rises from 88.0 ºC to its boiling point, what mass of water is in the sample?
A. 225 g
B. .255 g
C. 2.55 g
D. 25 g

Respuesta :

Answer:

c

Explanation:

heatmass*density*temp. difference=

Cricetus

In the sample, the mass of water is "225 g".

Given:

Heat energy,

  • [tex]Q = 1.13\times 10^4 \ J[/tex]

Specific heat,

  • [tex]c = 4.186 \ J/g.^{\circ} C[/tex]

The change in temperature,

  • [tex]\Delta T =boiling \ point - 88.0^{\circ} C[/tex]

              [tex]= 100-88[/tex]

              [tex]= 12 ^{\circ} C[/tex]

As we know,

→             [tex]Q = mc \Delta T[/tex]

→ [tex]1.13\times 10^4=m\times 4.186\times 12[/tex]        

→              [tex]m = \frac{1.13\times 10^4}{4.186\times 12}[/tex]

→                   [tex]= 224.956 \ g[/tex]

or,

→                   [tex]= 225 \ g[/tex]

Thus the answer above is appropriate.

     

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