Question

At 1 atm, how much energy is required to heat 0.0590 kg0.0590 kg of ice at −12.0 ∘C−12.0 ∘C to steam at 155.0 ∘C155.0 ∘C ?

STRATEGY:

1. Calculate the energy for each temperature change or phase change individually.
   1. The energy needed to heat 0.0590 kg0.0590 kg of ice from −12.0 ∘C−12.0 ∘C to its melting point.
   2. The energy needed to melt 0.0590 kg0.0590 kg of ice at its melting point.
   3. The energy needed to heat 0.0590 kg0.0590 kg of liquid water from the melting point to the boiling point.
   4. The energy needed to boil 0.0590 kg0.0590 kg of water at its boiling point.
   5. The energy needed to heat 0.0590 kg0.0590 kg of steam from the boiling point to 155.0 ∘C155.0 ∘C.
2. Sum the energies from each step and convert to kilojoules.

Step 1:

?A=??Δ?=(0.0590 kg)(2087Jkg⋅K)(12.0 ∘C)=1480 JqA=mcΔT=(0.0590 kg)(2087Jkg⋅K)(12.0 ∘C)=1480 J

?B=??f=(0.0590 kg)(3.336×105 J/kg)=19700 JqB=mLf=(0.0590 kg)(3.336×105 J/kg)=19700 J

?C=??Δ?=(0.0590 kg)(4190Jkg⋅K)(100 ∘C)=24700 JqC=mcΔT=(0.0590 kg)(4190Jkg⋅K)(100 ∘C)=24700 J

?D=??v=(0.0590 kg)(2.256×106 J/kg)=1.33×105 JqD=mLv=(0.0590 kg)(2.256×106 J/kg)=1.33×105 J

?E=??Δ?=(0.0590 kg)(2100Jkg⋅K)(55.0 ∘C)=6800 J

What is the sum of the individual energies?

Convert the sum from joules to kilojoules.

Answer 1