Asked by Anon

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 16.0 g, and its initial temperature is -11.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.0 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.
Answered by DrBob222
Do you have the heat of fusion? The specific heat of ice? The specific heat of H2O?
Answered by Anon
Enthalpy of fusion
333.6 J/g
6010. 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)
Answered by DrBob222
Heat absorbed by ice raising T from -11.4 to zero.
q1 = mass ice x specific heat ice x (Tf-Ti).
q1 = 16.0g x 2.087 J/g x 11.4 = ?J

q2 = heat to melt the ice to liquid at zero C.
q2 = mass x heat fusion ice.
q2 = 16.0g x 333.6 J/g = ?

q3 = heat to raise T of melted ice at zero C to 31.0 C.
q3 = mass H2O x specific heat H2O x (Tf-Ti)
q3 = 16.0g x 4.184 J/g*C x 31 =?

Total heat absorbed = q1 + q2 + q3 = ?
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