A) \[540\,\,cal/g\]
B) \[536\,\,cal/g\]
C) \[270\,\,cal/g\]
D) \[480\,\,cal/g\]
Correct Answer: A
Solution :
Heat required to raise the temperature of \[40\text{ }g\] of water from \[{{25}^{o}}C\] to \[{{54.3}^{o}}C,\] is equivalent to sum of heat required to condense the steam. \[\therefore \] Heat required to raise the temperature of water by \[{{t}^{o}}C\] is \[={{m}_{1}}c\,\Delta {{t}_{1}}\] ?.(i) where c is specific heat of water and m the mass. Heat required to condense steam \[={{m}_{2}}L+{{m}_{2}}c\Delta {{t}_{2}}\] ?..(ii) Equating Eqs. (i) and (ii), we get \[{{m}_{2}}L+{{m}_{2}}c\,\Delta {{t}_{2}}={{m}_{1}}c\Delta {{t}_{1}}\] Given, \[{{m}_{2}}=2g,\] \[\Delta {{t}_{2}}={{(100-54.3)}^{o}}C={{45.7}^{o}}C\] \[m=40g\] \[\Delta {{t}_{1}}={{(54.3-25)}^{o}}C={{19.3}^{o}}C,\] \[c=1\,cal/g\] \[\Rightarrow \] \[2\times L+2\times 1\times 45.7=40\times 1\times 29.3\] \[\Rightarrow \] \[2L+91.4=1172\] \[\Rightarrow \] \[2L=1080.6\] \[\Rightarrow \] \[L=540.3\,cal/g\]
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