Question

How much heat would be reqired to melt 10.0 g of ice at 0 oC, warm the resulting liquid to 100 oC, and change it to steam at 110 oC?

Answer 1

`7217` calories

Explanation:

We know latent heat of melting of ice is `80` calories/`g`

So,to convert `10g` of ice at `0^@C` to same amount of water at the same temperature, heat energy required would be `80*10=800` calories.

now,to take this water at `0^@C` to `100^@C` heat energy required will be `10*1*(100-0)=1000` calories (using,`H=ms d theta` where,`m` is the mass of water,`s` is specific heat,for water it is `1` C.G.S unit,and `d theta` is the change in temperature)

Now,we know,latent heat of vaporization of water is `537` calories/`g`

So,to convert water at `100^@C` to steam at `100^@C` heat energy required will be `537*10=5370` calories.

Now,to convert steam at `100^@C` to `110^@C`,heat energy required will be `10*0.47*(110-100)=47` calories (specific heat for steam is `0.47` C.G.S units)

So,for this entire process heat energy required will be `(800+1000+5370+47)=7217` calories