Question

Find the mass of 0oC ice that 10 gof 100oC steam will completely melt. Latent heat of fusion of ice is 80cal/g oCand heat of vaporization of water 540 cal/g oC ?

Answer 1

The right approach is "80 g".

Step-by-step explanation:

Given:

`L_f=80 \ Cal/g`

`L_v=540 \ Cal/g`

`S=1 \ Cal/g`

Now,

The amount of heat cooling will be:

=
`mL_v+mS \Delta T`

=
`(10* 540)+10* 1* (100-0)`

=
`5400+1000`

=
`6400 \ Cal`

then,

⇒
`m_(ice) L_f=6400`

`m_(ice)* 80=6400`

`m_(ice)=(6400)/(80)`

`=80 \ g`