How much heat (in kJ) is required to warm 12.0 g of ice, initially at -12.0 ∘C, to steam at 112.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C. kJ
There are many steps invoved in this
Step 1: to change ice at -12oC to 0oC
Q1= m x Cp x dT
Q1= 12.0 g (2.09 J/ g C) (0 - -12)
Q1= 300.96J
step2:to melt ice into liquid water
Q2 = mx Hf
Q2= 12.0 g (334 J/g) = 4008 J
Step 3: to warm the liquid water to 100 celsius:
Q3 = m xCp x dT
Q3= 12.0 grams (4.184 J/g-C)(100 C) = 5020.8J
Step 4: to evaporate the liquid water into steam
Q4= = m x Hvap
Q4 =12.0 grams (2260 J /g) = 27120J
step 5:to heat steam to 112C
Q5 = m x Cp x dT
Q5 =12.0 g (1.84 J/g-C) (112 - 100 C ) = 264.96 J
Total heat = Q1+Q2+Q3 +Q4 + Q5 = 36714.72 J = 36.7kJ
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