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A 58.1g sample of quartz is put into a calorimeter that contains 250.0g of water. The quartz sample starts off at 98.4°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 27.9°C. The pressure remains constant at 1 atm. Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

Respuesta :

Answer:

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Explanation:

Given:

  • mass of sample quartz, [tex]m_q=58.1\ g[/tex]
  • mass of water in calorie-meter, [tex]m_w=250\ g[/tex]
  • initial temperature of quartz, [tex]T_{qi}=98.4^{\circ}C[/tex]
  • initial temperature of water, [tex]T_{wi}=25^{\circ}C[/tex]
  • final temperature of the mixture, [tex]T_{f}=27.9^{\circ}C[/tex]

We have:

  • specific heat capacity of water, [tex]m_c=4.186\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.

Therefore,

heat gained by the water = heat lost by the quartz

[tex]Q_w=Q_q[/tex]

[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]

    where: [tex]c_q=[/tex] specific heat capacity of quartz

[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]