Answer:
Heat energy needed = 1243.45 kJ
Explanation:
We have
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C
Here wee need to convert 0.500 kg water from 50°C to vapor at 110°C
First the water changes to 100°C from 50°C , then it changes to steam and then its temperature increases from 100°C to 110°C.
Mass of water = 500 g
Heat energy required to change water temperature from 50°C to 100°C
[tex]H_1=mC\Delta T=500\times 4.18\times (100-50)=104.5kJ[/tex]
Heat energy required to change water from 100°C to steam at 100°C
[tex]H_2=mL=500\times 2257=1128.5kJ[/tex]
Heat energy required to change steam temperature from 100°C to 110°C
[tex]H_3=mC\Delta T=500\times 2.09\times (110-100)=10.45kJ[/tex]
Total heat energy required
[tex]H=H_1+H_2+H_3=104.5+1128.5+10.45=1243.45kJ[/tex]
Heat energy needed = 1243.45 kJ
