A) \[{{270}^{\text{o}}}C\]
B) \[{{230}^{\text{o}}}C\]
C) \[{{100}^{\text{o}}}C\]
D) \[{{50}^{\text{o}}}C\]
Correct Answer: C
Solution :
Heat required by 1 g ice at \[{{0}^{\text{o}}}C\] to melt into 1 g water at \[{{0}^{\text{o}}}C\], Q1 = mL (L = latent heat of fusion) \[=1\times 80\,\,=80\,cal\] (L = 80 cal/g ) Heat required by 1 g of water at \[{{0}^{\text{o}}}C\] to boil at \[{{100}^{\text{o}}}C\], \[{{Q}_{2}}=mc\Delta \theta \] (c = specific heat of water) \[=1\times 1\,\,(100-0)\] (c = 1 cal/g °C) = 100 cal Thus, total heat required by 1 g of ice to reach a temperature of \[{{100}^{\text{o}}}C\], \[Q={{Q}_{1}}+{{Q}_{2}}\] \[=80+100=180\,cal\] Heat available with 1g of steam to condense into 1 g of water at \[{{100}^{\text{o}}}C\], \[Q'=mL'\] (L' = latent heat of vaporisation) \[=1\times 536\,cal\] (L? = 536 cal/g) = 536 cal Obviously, the whole steam will not be condensed and ice will attain temperature of \[{{100}^{\text{o}}}C\]. Thus, the mixture of temperature is \[{{100}^{\text{o}}}C\].
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