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 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, \[m{{ & }_{2}}=2g,\] \[\Delta {{t}_{2}}={{(100-54.3)}^{o}}C={{45.7}^{o}}C\] \[{{m}_{1}}=40\,g\] \[\Delta {{t}_{1}}={{(54.3-25)}^{o}}C={{19.3}^{o}}C,\] \[c=1\,cal/g\] \[\Rightarrow \] \[2\times L+2\times 1\times 45.7=40\times 1\times 29.3\] \[\Rightarrow \] \[2L=91.4=1172\] \[\Rightarrow \] \[2L=1080.6\] \[\Rightarrow \] \[L=540.3\,cal/g\]
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