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

One kilogram of ice at $0°C$ is mixed with one kilogram of water at $80°C.$ The final temperature of the mixture is........ $^oC$

$($Take : specific heat of water$ = 4200\,J\,k{g^{ - 1}}\,{K^{ - 1}}$, latent heat of ice $ = 336\,kJ\,k{g^{ - 1}})$

A

$40$

B

$60$

C

$0$

D

$50$

Solution

(c) ${\theta _{{\rm{mix}}}} = \frac{{{m_W}{\theta _W} – \frac{{{m_{\,i}}{L_{\,i}}}}{{{c_W}}}}}{{{m_{\,i}} + {m_W}}}$

$\because \,{m_i} = {m_W}$ $\Rightarrow$ ${\theta _{mix}} = \frac{{{\theta _W} – \frac{{{L_i}}}{{{c_W}}}}}{2}$$ = \frac{{80 + 0 – \frac{{336}}{{4.2}}}}{2} = 0^\circ C$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.