To treat a burn on your hand, you decide to place an ice cube on the burned skin

Question

To treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 12.6 g, and its initial temperature is -14.5 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.4 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.

Constants may be found here.

Quantity per gram per mole Enthalpy of fusion 333.6 J/g 6010. J/mol Enthalpy of vaporization 2257 J/g 40660 J/mol Specific heat of solid H2O (ice) 2.087 J/(g·°C) * 37.60 J/(mol·°C) * Specific heat of liquid H2O (water) 4.184 J/(g·°C) * 75.37 J/(mol·°C) * Specific heat of gaseous H2O (steam) 2.000 J/(g·°C) * 36.03 J/(mol·°C) *

Explanation / Answer

here total 3- stages took place by ice to reach water

stage-1 , ice from -14.5 to 0 C

heat absorbed = m C (T2-T1) = 12.6 g x 2.087 J/g C x ( 0-(-14.5)) = 381.3 J

stage-2 :ice at 0 C to water at 0 C

heat absorbed = m x heat of fusion = 333.6 J/g x 12.6 g == 423.36 J

stage-3 : Water from 0 C to water 31.4 C

heat absorbed = m C (T2-T1) = 12.6 g x 4.184 J/g C x ( 31.4 - 0) = 1655.35 J

total heat = 1655.35+423.36+381.3 = 2460 J

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