Question

$1$g of ice at $0^o$C is converted into stream at ${100}^o$C. The amount of heat required to do so is.

• A. $80$cal
• B. $536$cal
• C. $716$cal
• D. $0.716$kcal

Answer 1

The correct option is C $716$cal

The correct answer is option(D).

Ice$\to$water$\to$water$\to$steam

0° ice $\to$ 0° water $\to$ 100° water $\to$ 100° steam

So from 0°C ice to 0°C water heat is $Q_1=m{L}_f$, i.e $L_f$=latent heat of fusion

From 0°C water $\to$ 100°C water $Q_2=mC\times (t_2-t_1)$

From 100°C water $\to$ 100°C steam $Q_3=m{L}_v$i.e $L_v$=latent heat of vaporisation

Total heat,

$Q=Q_1+Q_2+Q_3$

$Q=m[L_f\times C(t_2-t_1)\times L_v]$

$Q=1[80+1(100-0)+536]$

$Q=716cal=0.716kcal$