Question

Work done in converting one gram of ice at $-{10}^{\circ }\text{C}$ into steam at ${100}^{\circ }\text{C}$ is -
[Given, specific heat of ice $c_i=0.5{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1}$, specific heat of water $c_w=1{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1}$, latent heat of fusion of ice $L_f=80\text{cal/g}$, & latent heat of vaporization of steam $L_v=540\text{cal/g}$]

• A. $725\text{J}$
• B. $3045\text{J}$
• C. $5023\text{J}$
• D. $1128\text{J}$

Answer 1

The correct option is B $3045\text{J}$

Pictorial representation of conversion of ice at $-{10}^{\circ }\text{C}$ into steam at ${100}^{\circ }\text{C}$ is given below
($c_i=$ specific heat of ice, $c_w=$specific heat of water)


Let $Q$ be the total heat required to convert ice into steam.
From the figure, we can deduce that,
$Q=Q_1+Q_2+Q_3+Q_4$
$Q=m{c}_i\Delta \theta +m{L}_f+m{c}_w\Delta \theta +m{L}_v$
Substituting given values,
$\Rightarrow Q=1\times 0.5\times [0-(-10\left)\right]+1\times 80+1\times 1\times (100-0)+1\times 540$
$\therefore Q=725\text{cal}$
Hence, work done in converting ice (at $-{10}^{\circ }\text{C}$) into steam (at ${100}^{\circ }\text{C}$) is given by,
$W=JQ=4.2\times 725=3045\text{J}$
Thus, option (b) is the correct answer.