Exactly 20.0 mL of water at 32.0 °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 1.15 kg. What is the change in temperature of the skillet?

98°C

Explanation:

Given dа ta:

Volume of water = 20.0 mL

Temperature = 32°C

Mass of skillet = 1.15 kg

Change in temperature = ?

Solution:

Q = m. c.ΔT

ΔT = 100°C - 32°C

ΔT = 68°C

Q = m. c.ΔT

Q = 20 g * 4.184 j/g.°C * 68°C

Q = 5690.24 j

Heat of vaporization of water = 2257 j/g

q = 2257 j/g * 20 g

q = 45140 j

Total heat gained by water = 5690.24 j + 45140 j

Total heat gained by water = 50830.24 j

Number of moles of iron:

Number of moles = mass / molar mass

Number of moles = 1150/55.85

Number of moles = 20.59 mol

Molar heat capacity of iron is = 25.19 j/mol.°C

50830.24 j / 25.19 j/mol.°C/20.59 mol

98°C