• # question_answer If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is:           [AIPMT 1999] A)        ${{270}^{o}}C$        B)       ${{230}^{o}}C$ C)  ${{100}^{o}}C$        D)       ${{50}^{o}}C$

 Heat required by 1 g ice at ${{0}^{o}}C$ to melt into 1g water at ${{0}^{o}}C$, ${{Q}_{1}}=mL$       (L = latent heat of fusion) $=1\times 80=80\text{ cal}$      $(L=80\text{ }cal/g)$ Heat required by 1 g of water at ${{0}^{o}}C$ to boil at ${{100}^{o}}C$, ${{Q}_{2}}=mc\Delta \theta$  (c = specific heat of water) $=1\times 1\text{ }(100-0)$        $(c=1\text{ }cal/g{}^\circ C)$ $=100\text{ }cal$ Thus, total heat required by 1 g of ice to reach a temperature of ${{100}^{o}}C$, $Q={{Q}_{1}}+{{Q}_{2}}$ $=80+100=180\,cal$ Heat available with 1g of steam to condense into 1g of water at ${{100}^{o}}C$, $Q'=mL'$         (L' = latent heat of vaporisation) $=1\times 536\text{ }cal$    (L = 536 cal/g) = 536 cal Obviously, the whole steam will not be condensed and ice will attain temperature of ${{100}^{o}}C$. Thus, the mixture of temperature is ${{100}^{o}}C$.