A) \[{{100}^{o}}C\]
B) \[{{55}^{o}}C\]
C) \[{{0}^{o}}C\]
D) \[{{50}^{o}}C\]
Correct Answer: A
Solution :
: Here, Mass of ice, \[{{m}_{ice}}=1g\] Mass of steam, \[{{m}_{steam}}=1g\] Latent heat of fusion of ice, \[{{L}_{ice}}=80\,ca\operatorname{l}\,\,{{g}^{-1}}\] Latent heat of steam, \[{{L}_{steam}}=540\,ca\operatorname{l}\,\,{{g}^{-1}}\] Specific heat of water, \[{{S}_{water}}=1\,cal\,{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}\] Heat required to convert ice at 0°C to water at \[{{100}^{o}}C\]is \[{{Q}_{1}}={{m}_{ice}}\,{{L}_{ice}}+{{m}_{ice}}\,{{S}_{water}}\Delta T\] \[=(1g)\,(80\,cal\,{{g}^{-1}})\] \[+(1g)(1cal\,{{g}^{-1}}{{\,}^{o}}{{C}^{-1}})({{100}^{o}}C-{{0}^{o}}C)\] \[=80cal+100cal=180cal\] Heat released by steam at \[{{100}^{o}}C\] to condense into water at \[{{100}^{o}}C\] is \[{{Q}_{2}}={{m}_{steam}}{{L}_{steam}}\] \[=(1g)(540\,cal\,{{g}^{-1}})=540cal\] As \[{{Q}_{1}}<{{Q}_{2}}\], whole of the steam will not condense, so the temperature of the mixture is \[{{100}^{o}}C\].
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