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 10.4 g, and its initial temperature is -12.1 °C. The water resulting from the melted ice reaches the temperature of your skin, 29.7 °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

The ice undergoes folllowing changes

i) ice reaching temp from -12.1 to Zero deg Cel

ii) Then it melts at Zero Deg Cel

iii) Then it raising temperature from Zero to 29.7 Deg cel

Q = mcT

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kgK), is a symbol meaning "the change in"

T = change in temperature (Kelvins, K)

Q = 10.4 x 2.087 x 12.1 + 10.4 x 333.6 + 10.4 x 29.7 x 4.184

Q = 7422.73264 Joules

Hence 7422.73 Joules or 7.422 Kilo joules of heat absorbed by ice

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