A) \[{{50}^{o}}C\]
B) \[{{100}^{o}}C\]
C) \[{{67}^{o}}C\]
D) \[{{33}^{o}}C\]
Correct Answer: B
Solution :
Heat taken by ice to raise its temperature to \[{{100}^{o}}C\]is \[Q=mL+mc\Delta \theta \] where \[L\] is latent heat, \[c\] is specific heat, and \[\Delta \theta \] is temperature variation. Given,\[m=1\,\,g,\,\,L=80\,\,cal/g\] \[c=1\,\,cal/g,\,\,\Delta \theta ={{100}^{o}}C\] \[{{Q}_{1}}=1\times 80+1\times 1\times 100=180\,\,cal\] Heat given by steam when condensed is \[{{Q}_{2}}={{m}_{2}}{{L}_{2}}=1\times 540=540\,\,cal\] As \[{{Q}_{2}}>{{Q}_{1}}\], hence temperature of mixture will remain\[{{100}^{o}}C\].
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