Norman Manchu
Last Activity: 6 Years ago
m=1gram
The change takes place as-
-10°C(ice)---0°C(ice)---0°C(water)---100°C(water)---100°C(steam)
Tip:When state changes temperature doesn't Use energy=ms?(theta) when temperature changes and energy=mL when the state changes
-10°C(ice)to 0°C(water)=ms?(theta)=5cal
0°C(ice)to 0°C(water)=mL=80cal
0°C(water)to 100°C(water)=100cal
100°C(water)to 100°C(steam)=540cal
Total=725cal=3045joules
Note:1cal=4•2joules|||
Latent heat of ice to water=80cal
Latent heat of water to stean=540cal
Specific heat of ice=0•5cal-per-gram
Specific heat of water=1cal-per-gram
Done|