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 data:

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

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