Question

3rd attempt Feedback See Periodic Table See Hint Exactly 10.5 mL of water at 30.0 °C is added to a hot iron skillet. All of the water is converted to steam at 100.0°C. The mass of the skillet is 1.50 kg. What is the change in temperature of the skillet? 47.82 C 2nd attempt > 1st attempt

Answer 1

Temperature change of iron = -40.2 ^oC

Expanation

volume of water = 10.5 mL

mass of water = (volume of water) * (density of water)

mass of water = (10.5 mL) * (1.00 g/mL)

mass of water = 10.5 g

Process 1 : All water is heated from 30.0 ^oC to 100 ^oC

H1 = (mass of water) * (specific heat water) * (final temperature - initial temperature)

H1 = (10.5 g) * (4.184 /g.^oC) * (100 ^oC - 30.0 ^oC)

H1 = 3075.24 J

Process 2 : All water is converted to steam

H2 = (mass of water) * (heat of vaporization)

H2 = (10.5) * (2256.42 J/g)

H2 = 23692.4 J

Total heat gained by water = H1 + H2

Total heat gained by water = 3075.24 J + 23692.4 J

Total heat gained by water = 26767.6 J

Total heat gained by water = 26.77 kJ

Total heat lost by iron = -Total heat gained by water

Total heat lost by iron = -26.77 kJ

Temperature change of iron = (Total heat lost by iron) / [(mass of iron) * (specific heat iron)]

Temperature change of iron = (-26.77 kJ) / [(1.50 kg) * (0.444 kJ/kg.^oC)]

Temperature change of iron = -40.2 ^oC