Heat and Thermodynamics 1 Question 1
1. When $M _1$ gram of ice at $-10^{\circ} C$ (specific heat $=0.5 cal$ $g^{-1}{ }^{\circ} C^{-1}$ ) is added to $M _2$ gram of water at $50^{\circ} C$, finally no ice is left and the water is at $0^{\circ} C$. The value of latent heat of ice, in cal $g^{-1}$ is
(a) $\frac{50 M _2}{M _1}-5$
(b) $\frac{50 M _1}{M _2}-50$
(c) $\frac{50 M _2}{M _1}$
(d) $\frac{5 M _2}{M _1}-5$
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Solution:
- Key Idea In such kind of heat transfer problems, Heat given by water $=$ Heat gained by ice and (heat) $) _{\text {liquid }}=$ mass $\times$ specific heat $\times$ (Heat) $) _{\text {solid }}=($ mass $\times$ latent heat $)$ temperature $={m _l s _l \Delta T _l }$
- (mass $\times$ specific heat $\times$ temperature) $=\left(m _s \times L\right)+m _s \times s _s \times \Delta T _s$
Heat given by water is (specific heat of water is $1 cal g^{-1}{ }^{\circ} C^{-1}$ )
$$ (\Delta H) _{\text {water }}=M _2 \times 1 \times(50-0)=50 M _2 $$
Heat taken by ice is
$$ \begin{aligned} & (\Delta H) _{\text {ice }}=M _1 \times 0.5 \times[0-(-10)]+M _1 \times L _{\text {ice }} \\ \Rightarrow \quad(\Delta H) _{\text {ice }} & =5 M _1+M _1 L _{\text {ice }} \end{aligned} $$
Comparing Eqs. (i) and (ii), we get
$$ \begin{aligned} \therefore \quad(\Delta H) _{\text {water }} & =(\Delta H) _{\text {ice }} \\ 50 M _2 & =5 M _1+M _1 L _{\text {ice }} \\ \Rightarrow \quad L _{\text {ice }} & =\frac{50 M _2-5 M _1}{M _1} \\ L _{\text {ice }} & =50 \frac{M _2}{M _1}-5 \end{aligned} $$
