Question

The amount of heat required to convert $5\text{g}$ of ice at $0^{\circ }\text{C}$ to $5\text{g}$ of steam at ${100}^{\circ }\text{C}$ is -
[Latent heat of vaporization and fusion are $L_v=540{\text{cal g}}^{-1}$ and $L_f=80{\text{cal g}}^{-1}$]

• A. $3100\text{cal}$
• B. $3200\text{cal}$
• C. $3600\text{cal}$
• D. $4200\text{cal}$

Answer 1

The correct option is C $3600\text{cal}$

Pictorial representation of conversion of ice at $\left(0^{\circ }\text{C}\right)$ into steam at $\left({100}^{\circ }\text{C}\right)$ is shown below.


Given,
Mass of ice $\left(m_i\right)=5\text{g}$
Temperature of ice $\left(T_1\right)=0^{\circ }\text{C}$
Temperature of steam $\left(T_2\right)={100}^{\circ }\text{C}$
Latent heat of vaporization $\left(L_v\right)=540{\text{cal g}}^{-1}$
Latent heat of fusion $\left(L_f\right)=80{\text{cal g}}^{-1}$

Since there are no other sources and sinks which can change the amount of substance used in the problem, we can say that mass of steam = mass of water = initial mass of the ice.
i.e $m_i=m_w=m_s$
Total heat required during the phase change is
$Q=Q_1+Q_2+Q_3$
$Q=m_i\times L_f+m_w\times c_w\times (T_2-T_1)+m_w\times L_v$

$\therefore Q=5\times 80+5\times 1\times (100-0)+5\times 540$
$=3600\text{cal}$
Thus, option (c) is the correct answer.