MCQ
$300\, gm$ of water at $25°C$ is added to $100\, gm$ of ice at $0°C$. The final temperature of the mixture is........ $^oC$
- A$ - \frac{5}{3}$
- B$ - \frac{5}{2}$
- C$-5$
- ✓$0$
✓
Answer
Correct option: D.
$0$
d
(d) ${\theta _{{\rm{mix}}}} = \frac{{{m_W}{\theta _W} - \frac{{{m_{\,i}}{L_{\,i}}}}{{{S_W}}}}}{{{m_{\,i}} + {m_W}}}$$ = \frac{{300 \times 25 - \frac{{100 \times 80}}{1}}}{{100 + 300}} = - {1.25^o}C$
(d) ${\theta _{{\rm{mix}}}} = \frac{{{m_W}{\theta _W} - \frac{{{m_{\,i}}{L_{\,i}}}}{{{S_W}}}}}{{{m_{\,i}} + {m_W}}}$$ = \frac{{300 \times 25 - \frac{{100 \times 80}}{1}}}{{100 + 300}} = - {1.25^o}C$
Which is not possible. Hence ${\theta _{mix}} = {0^o}C$
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