Question

Exactly 10.1 mL of water at 31.0 degree C are added to a hot iron skillet. All of the water is converted into steam at 100.0 degree C. The mass of the pan is 1.50 kg and the molar heat capacity of iron is 25.19 J/(mol middot degree C). What is the temperature change of the skillet? 104 degree C

Answer 1

Heat lose of iron           = heat gain of water

mc $\Delta$ t                           = mc $\Delta$ t

1500*0.45*(t-100)         = 10.1*4.18*(100-31)

     t   = 104.3⁰C

temperature change of the skillet = 104.3-100 = 4.3⁰c