Question

What is the amount of heat required (in calories) to convert $10 \, g$ of ice at $-10 \, ℃$ into steam at $100 \, ℃$ ? Given that latent heat of vaporization of water is $540 \, cal \, g^{- 1}$ , latent heat of fusion of ice is $80 \, cal \, g^{- 1}$ , the specific heat capacity of water and ice are $1 \, cal \, g^{- 1} \,{}^\circ C^{- 1}$ and $0.5 \, cal \, g^{- 1} \,{}^\circ C^{- 1}$ respectively.

Answer 1

10 g of ice at $-10℃$ to ice at 0 $℃$
$Q_{1}=cm, \, \Delta \theta =0.5\times 10\times 10=50$ cal
10 g of ice 0 $℃$ to water at 0 $℃$
$Q_{2}=mL=10\times 80=800 \, $ cal
10 g of water at 0 $℃$ to water at 100 $℃$
$Q_{3}=cm, \, \Delta \theta =1\times 10\times 100=1000$ cal
10 g water at 100 $℃$ to steam at 100 $℃$
$Q_{4}=mL=10\times 540=5400$ cal
Total heat required, $Q=Q_{1}+Q_{2}+Q_{3}+Q_{4} \, $
= 50+800+1000+5400 = 7250 cal