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 \[\text{100}{{\,}^{\text{o}}}\text{C}\]is \[Q=mL+mc\Delta \theta \] where L is latent heat, c is specific heat, and\[\Delta \theta \] is temperature variation. Given, \[m=1\,g,L=80\,cal/g,\] \[c=1\,cal/g,\,\Delta \theta =100{{\,}^{o}}C\] \[{{Q}_{1}}=1\times 80+1\times 1\times 100=180\,\text{cal}\] Heat given by steam when condensed is \[{{Q}_{2}}={{m}_{2}}{{L}_{2}}=1\times 540=540\,\text{cal}\] As \[{{\text{Q}}_{\text{2}}}\text{}\,{{\text{Q}}_{\text{1}}}\text{,}\]hence temperature of mixture will remain \[\text{100}{{\,}^{\text{o}}}\text{C}\text{.}\]
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