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 15.1 g, and its initial temperature is -12.2 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.3 °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.

Answer 1

specific heat of ice 2.01J/g-c⁰

heat of fustion     334J/g

specific heat of water = 4.184J/g-C⁰

q1   = mcT

          = 15.1*2.01*(0-(-12.2)

          = 15.1*2.01*12.2   = 370.3J to warm the ice to 0C⁰

q2      = 15.1*334         = 5043.4J fustion of ice
q3      = mct

          = 15.1*4.184*(28.3-0)   = 1785.95J to warm the melted ice to skine temperature

Total heat = q1 + q2 + q3

                   = 370.3+5043.4+ 1785.95 = 7199.65J


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