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How many grams of ice at 0 degrees C will be melted in bringing down a 62 kg patient's fever from 40 degrees C to 39 degrees C?

Answer :

Answer:

775 grams of ice

Explanation:

Heat loss by patient = heat gained by ice

Heat loss by patient = mass×specific heat capacity×(initial temperature - final temperature) = 62×4200×(40-39) = 260,400J

Heat gained by ice = mass × specific latent heat of fusion of ice = m×336000 = 336000m J

336000m = 260,400

m = 260400/336000 = 0.775kg = 0.775×1000 grams = 775 grams of ice

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Rewritten by : Barada

"640 g" of ice at 0°C will be melted.

Given:

  • Mass, m = 62 kg

We know,

  • Specific heat of human body = 3470 J/kg.C
  • Specific heat of ice = 2108 J/kg.C
  • Ice's heat of fusion, [tex]L_f[/tex] = 334000 J/kg

As we know,

→ [tex]Heat \ gain \ by \ ice = Heat \ loss \ by \ patient[/tex]

or,

→ [tex]mC \Delta t+m L_f = mC \Delta t[/tex]

By taking common,

→ [tex]m(C \Delta t+ L_f) = m C \Delta t[/tex]

By substituting the values, we get

→ [tex]m(2108\times 1+334000)= 62\times 3470\times 1[/tex]

[tex]m = 0.640 \ kg[/tex]

or,

[tex]m = 640 \ g[/tex]

Thus the above approach is correct.

Learn more about specific heat here:

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