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Ice in a freezer is at $-7^{\circ} C .100 \,g$ of this ice is mixed with $200 \,g$ of water at $15^{\circ} C$. Take the freezing temperature of water to be $0^{\circ} C$, the specific heat of ice equal to $2.2 \,J / g { }^{\circ} C$, specific heat of water equal to $4.2 \,J / g ^{\circ} C$ and the latent heat of ice equal to $335 \,J / g$. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to .......... $g$
$88$
$67$
$54$
$45$
Solution
(b)
Let $m$ gram of ice melt and water reaches $0^{\circ} C$.
Then,
Heat lost by ice $=$ Heat gained by water
$\Rightarrow m_{\text {ice }} L_{\text {ice }}+m_{\text {ice }} c_{\text {ice }} \Delta T_{\text {ice }}=m_{l v} c_v \Delta T_w$
$\Rightarrow m(335)+m(2.2)(0-(-7))$
$=200(42)(15-0)$
$\Rightarrow m(335+154)=12600$
$\Rightarrow m=\frac{12600}{3504} \approx 36 \,g$
So, ice left in mixture is $100-36 \approx 64 \,g$.
Hence, nearest option is $67 \,g$.
