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 18.8 g, and its initial temperature is -11.8 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.4 °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 below.

Heat Transfer Heat-transfer constants for H20 at 1 atm per gram 333.6 J/g 2257 J/g per mole Quantity Enthalpy of fusion Enthalpy of vaporization Specific heat of solid H20 (ice) Specific heat of liquid H20 (water) 4.184 J/(g.°C)*75.37 J/(mol.°C) * Specific heat of gaseous H20 (steam) 2.000 J/(g.°C) 36.03 /(mol.C)* 6010. J/mol 40660 J/mol Specific heats change slightly with temperature. The values given in this table are a compromise among the accepted averages from several top-selling general chemistry textbooks. The slight differences in these values will not significantly affect your answers on homework problems.

Explanation / Answer

Heat absorbed by ice raising T from -11.8 to zero.
q1 = mass ice x specific heat ice x (Tf-Ti).
q1 = 18.8g x 2.087 J/g x 11.8 = 462.98 J

q2 = heat to melt the ice to liquid at zero C.
q2 = mass x heat fusion ice.
q2 = 18.8g x 333.6 J/g = 6271.68 J

q3 = heat to raise T of melted ice at zero C to 28.4 C.
q3 = mass H2O x specific heat H2O x (Tf-Ti)
q3 = 18.8g x 4.184 J/g*C x 28.4 = 2233.93

Total heat absorbed = q1 + q2 + q3 = 8968.6 J ------------ answer

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