A) \[{{270}^{o}}C\]
B) \[{{230}^{o}}C\]
C) \[{{100}^{o}}C\]
D) \[{{50}^{o}}C\]
Correct Answer: C
Solution :
| 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 \[{{100}^{o}}C\]. |
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