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Physics Question on Calorimetry

One kilogram of ice at 0∘C0^{\circ} C is mixed with one kilogram of water at 80∘C80^{\circ} C. The final temperature of the mixture is (Take specific heat of water =4200 kJ/kg−∘C=4200\, kJ / kg -{ }^{\circ} C, Latent heat of ice =336 kJ/kg=336\, kJ / kg )

A

0∘C0^{\circ} C

B

40∘C40^{\circ} C

C

50∘C50^{\circ}C

D

60∘C60^{\circ}C

Answer

0∘C0^{\circ} C

Explanation

Solution

It is known that For water and ice mixing
θmix=mWθW−miLiCWmi+mW\theta_{mix}=\frac{m_{W} \theta_{W}-\frac{m_{i} L_{i}}{C_{W}}}{m_{i}+m_{W}}
θmix⁡=mWθW−miLicWmi+mW\theta_{\operatorname{mix}}=\frac{m_{W} \theta_{W}-\frac{m_{i} L_{i}}{c_{W}}}{m_{i}+m_{W}}
∵mi=mw\because m _{ i }= m _{ w }
⇒θmix=θW−LiCW2\Rightarrow \theta_{mix}=\frac{\theta_{ W }-\frac{ L _{ i }}{ CW }}{2}
=80−3364.22=0∘C=\frac{80-\frac{336}{4.2}}{2}=0^{\circ} C