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An ice bag containing 222 g of ice at 0 ∘C was used to treat sore muscles. When the bag was removed, the ice had melted and the liquid water had a temperature of 18 ∘C.
How many kilojoules of heat were absorbed?

Respuesta :

Answer:

90.9312 kJ

Explanation:

Given:

[tex]mass \ of\ ice\ (m)=222\ g=0.222\ kg[/tex]

[tex]initial\ temperature\ (T_o)=0^oC[/tex]

[tex]final\ temperature\ (T)=18^oC[/tex]

[tex]latent\ heat\ of\ ice\ to\ water\ (L)=334\ kJ/kg[/tex]

[tex]specific\ heat\ of\ water\ (c)=4.2\ kJ/kg^oC[/tex]

The ice absorbed the heat in 2 process:

  1. Ice melted into water at 0°C
  2. Water temperature went up from 0°C to 18°C

∑Q = Q melting + Q water

     [tex]=mL+mc(T-T_o)[/tex]

     [tex]=0.222(334)+0.222(4.2)(18-0)[/tex]

     [tex]=74.148+16.7832[/tex]

     [tex]=90.9312\ kJ[/tex]

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