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

Answer:

The right approach is "80 g".

Explanation:

Given:

$L_f=80 \ Cal/g$

$L_v=540 \ Cal/g$

$S=1 \ Cal/g$

Now,

The amount of heat cooling will be:

= $mL_v+mS \Delta T$

= $(10\times 540)+10\times 1\times (100-0)$

= $5400+1000$

= $6400 \ Cal$

then,

⇒ $m_{ice} L_f=6400$

$m_{ice}\times 80=6400$

        $m_{ice}=\frac{6400}{80}$

                $=80 \ g$