calculate the required heat energy to change 10g of ice at 0 degrees Celsius water vapour at 100 degrees Celsius
Answers
Answer:
Total heat is required to change 10 g ice at 0°C to steam at 100°C is 7200 cal.
Explanation:
Latent heat of fusion of ice= \Delta H_{fus}=80cal/gΔH
fus
=80cal/g
Heat required to melt 10 g of ice = Q
Q=m\times \Delta H_{fus}Q=m×ΔH
fus
=10 g\times 80cal/g=800 cal=10g×80cal/g=800cal
Heat require to raise the temperature of water from 0°C to 100°C = Q'
Specific heat of water =c= 1 cal/g°C
ΔT = 100°C - 0°C = 100°C
Q'=mc\times \Delta TQ
′
=mc×ΔT
=10 g\times 1 cal/g^oC\times 100^oC=1000 cal=10g×1cal/g
o
C×100
o
C=1000cal
Latent heat of vaporization of water= \Delta H_{vap}=540cal/gΔH
vap
=540cal/g
Heat required to vaporize 10 g of water = Q''
Q''=m\times \Delta H_{vap}Q
′′
=m×ΔH
vap
=10 g\times 540 cal/g=5400 cal=10g×540cal/g=5400cal
Total heat is required to change 10 g ice at 0°C to steam at 100°C.
=Q+Q'+Q''=800 cal+1000 cal+5400 cal = 7200 cal=Q+Q
′
+Q
′′
=800cal+1000cal+5400cal=7200cal