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.
3 answers
Do you have the heat of fusion? The specific heat of ice? The specific heat of H2O?
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)
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)
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 = ?
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 = ?
3 answers