Question

Exactly 18.6 mL of water at 35.0 °C are added to a hot iron skillet. All of the water is converted into steam at 100.0°C. The mass of the pan is 1.50 kg and the molar heat capacity of iron is 25.19 J/(mol·°C). What is the temperature change of the skillet?

Answer 1

According to first law of thermodynamics ;

energy is conserved

heat lost by skillet = heat gain by water

water gain heat to become saturated steam ie. from 35 ^oC water to 100 ^oC steam

heating of water will occur in two steps

1. water from 35 ^oC to 100 ^oC sensible heat

2. water to vapour at 100 ^oC latent heat

step 1.

density of water = 1g/mL

mass = density x volume = 1g/ml x 18.6 mL = 18.6g

specific heat capacity of water Cpw = 4.18 J/mol.^oC

∆T = 100^oC - 35 ^oC = 65 ^oC

q1 = mCpw∆T = 18.6g x 4.18J/^oC.g x 65 ^oC = 5053.62 J

2nd step in heating ie. vaporization

∆H_vap = 2258.33 J/g

q2 = m∆H_vap = 18.6g x 2258.33 J/g = 42005 J

total heat absorbed by water = q1 + q2 = 47058.62 J

this heat is used to cool iron

miCpi∆T = 47058.62 J

1.5kg = 1500g

we have molar specific heat ,but here we are dealing in gram specific heat

we have to convert it to gram specific heat to work with

molar mass of Iron = 55.845 g per 1 mole

Cp = 25.19 J/mol.^oC = 25.19J/55.845g .^oC = 0.451 J/g.^oC

1500g x 0.451J/g.^oC x ∆T = 47058.62 J

∆T = 69.56 ^oC Ans