How much heat Q is required to change 17.0 g of ice at − 6.00 ∘...

How much heat Q is required to change 17.0 g of ice at − 6.00 ∘ C to 17.0 g of steam at 137 ∘ C ?

Solutions

Expert Solution

Ti = -6.0 oC
Tf = 137.0 oC
here
Cs = 2.09 J/goC

Heat required to convert solid from -6.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 17 g * 2.09 J/goC *(0--6) oC
= 213.18 J

Hf = 333.6 J/g

Heat required to convert solid to liquid at 0.0 oC
Q2 = m*Lf
= 17.0g *333.6 J/g
= 5671.2 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = m*Cl*(Tf-Ti)
= 17 g * 4.184 J/goC *(100-0) oC
= 7112.8 J

Hv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC
Q4 = m*Lv
= 17.0g *2260.0 J/g
= 38420 J

Cg = 2.01 J/goC

Heat required to convert vapour from 100.0 oC to 137.0 oC
Q5 = m*Cg*(Tf-Ti)
= 17 g * 2.01 J/goC *(137-100) oC
= 1264.29 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 213.18 J + 5671.2 J + 7112.8 J + 38420 J + 1264.29 J
= 52681 J

= 52.7 KJ
Answer: 52.7 KJ

queen_honey_blossom answered 1 year ago

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