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A 40.0-g block of ice at -15.00°C is dropped into a calorimeter (of negligible heat capacity) containing water at 15.00°C.

When equilibrium is reached, the final temperature is 8.00°C.

How much water did the calorimeter contain initially?

The specific heat of ice is 2090 J/kg • K, that of water is 4186 J/kg • K, and the latent heat of fusion of water is 33.5 × 104 J/kg.

Respuesta :

Answer:

[tex]m_w=545.817\ g[/tex]

Explanation:

Given:

  • mass of ice block, [tex]m_i=40\ g[/tex]
  • initial temperature of ice block, [tex]T_i_i=-15^{\circ}C[/tex]
  • initial temperature of water, [tex]T_i_w=15^{\circ}C[/tex]
  • final temperature of mixture, [tex]T_f=8^{\circ}C[/tex]
  • specific heat of ice, [tex]c_i=2.09\ J.g^{-1}[/tex]
  • specific heat of water, [tex]c_w=4.186\ J.g^{-1}[/tex]
  • Latent heat of fusion of water, [tex]L=335\ J.g^{-1}[/tex]

Now, assuming that there is no heat loss out of the mixture:

⇒ heat absorbed by the ice = heat rejected by the water

[tex]m_i(c_i\times \Delta T_i+L+c_w\times \Delta T_w)=m_w.c_w.\Delta T_w[/tex]

[tex]40(2.09\times (0-(-15))+335+4.186\times (8-0))=m_w\times 4.186\times (15-8)[/tex]

[tex]m_w=545.817\ g[/tex]

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