Gujarati
10-1.Thermometry, Thermal Expansion and Calorimetry
medium

$300\, gm$ 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

$-5$

D

$0$

Solution

(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 ${\theta _{mix}} = {0^o}C$

Standard 11
Physics

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