How much heat is needed to change 20 grams of ice at 0 celsius to steam at 100 celsius??

Question

How much heat is needed to change 20 grams of ice at 0 celsius to steam at 100 celsius??

Answer 1

this Q. requires the involvement of latent heat of fusion. specific heat capacity and latent heat of vaporization

heat required to change 20 g of ice into 20 g of water= mass * latent heat of fusion

= 20 g * 80 cal/g

=1600 cal=1.6 k cal

heat required to change the temp. from 0 celsius to 100 celsius

=mass*specific heat capacity*change in temp.

=20g * 1cal/g*celsius * 100celsius

=2000cal=2 k cal

heat required to change water to vapour

=mass * latent heat of vaporization

=20 g * 540cal/g

=10800cal=10.8 k cal

total heat required

=(1.6+2+10.8)k cal

=14.4 k cal

Answer 2

This looks like a homework problem so I won't simply give you the answer but I can explain....

Multiply the mass of the ice by the latent heat of fusion-this will give you the amount of heat needed to turn the ice to liquid water at zero degrees.

Now multiply the mass of water with the specific heat of water. This is the amount of heat needed to raise the temperature of the water by 1 degree. So you need to multiply this by the total temperature change, ie 100.

You now have water sitting at 100 degrees. To find the heat needed to turn the water to steam at 100 degrees multiply the mass of the water by the latent heat of fusion.

Add all three of these numbers up and you have the answer.

Good luck.

Answer 3

Ahhhhhh........100 Celsius