heat required to convert 1g of ice at 0 degree celcius into steam at 100 degree celcius is
Answers
Total heat required to convert 1 g of ice at 0°C into steam at 100°C is 716 cal.
Explanation:
Given Data:
Mass of ice=1 g
Temperature of ice = 0 °C
- The ice at 0°C is needed to convert to water at the same temperature i.e. 0°C
Heat required at this stage
= Mass of the ice x Latent heat of fusion of ice
=1 x 80 = 80 cal
- Now increase the water temperature from 0°C to 100°C by using the formula.
Heat required = Mass of water x rise in temperature x specific heat of water
= 1 x 100 x 1 = 100 cal
- Now convert water into vapour state at 100°C
Heat required for this
= Mass of water x Latent heat
= 1 x 536 =536 cal
Total heat required
=80 +100 +536 = 716 cal
Answer:
Q = 1 × 80 + 1 × 1 × 100 + 1 × 540 = 720 cal
Explanation:
Mass of ice=1 g
Temperature of ice = 0 °C
The ice at 0°C is needed to convert to water at the same temperature i.e. 0°C
Heat required at this stage
= Mass of the ice x Latent heat of fusion of ice
=1 x 80 = 80 cal
Now increase the water temperature from 0°C to 100°C by using the formula.
Heat required = Mass of water x rise in temperature x specific heat of water
= 1 x 100 x 1 = 100 cal
Now convert water into vapour state at 100°C
Heat required for this
= Mass of water x Latent heat
= 1 x 540 = 540 cal
Total heat required
=80 +100 +540 = 720 cal
THIS ANSWER IS FOR 1g OF STEAM!!!!!!!