Question

How much heat is required to convert `10g` of ice at `-10^@C` into steam at `100^@C`?

Answer 1

`7300` calories

Explanation:

When `10g` of ice at `-20°C` s being converted into steam at `100°C`, there are four stages.

1. Ice at `-20°C` to ice at `0°C` - here it continues to be in the same state i.e. ice and hence heat required is `"mass"xx"specific heat"xx"change in temperature"` Specific heat for ice is `0.5` cal/g-°C.
2. Then from ice at `0°C` to water at `0°C` and heat required is `"mass"xx"latent heat"`, this latent heat required is for conversion of each unit mass of substance. From ice to water it is `80"cal per gram"`
3. Water at `0°C` to water at `100°C` - here it continues to be in the same state i.e. water and hence heat required is `"mass"xx"specific heat"xx"change in temperature"` Specific heat for water is `1` cal/g-°C.
4. Then from water at `100°C` to steam at `100°C` and heat required is `"mass"xx"latent heat"`, this latent heat required is for conversion of each unit mass of substance. From water to steam, it is `540"cal per gram"`

Hence, to find heat is required to convert `10g` of ice at `-20°C` into steam at `100°C`,

first calculate heat required to convert `10g` of ice at `-20°C` to ice at `0°C`. As specific heat of ice is `0.5"cal/g°C"`, this is

`10xx20xx0.5=100` cal.

heat required to convert `10g` of ice to `10g` of water at `0°C` is

`10xx80=800` cal - as latent heat is `80"cal/g"`

heat required to convert `10g` of water at `0°C` to `10g` of water at `100°C` is

`10xx100x1=1000` cal

heat required to convert `10g` of water at `1000°C` to `10g` of steam at `100°C` is

`10xx540=5400` cal - as latent heat is `540"cal/g"`

Hence, total heat required is `100+800+1000+5400=7300` calories