InterviewSolution
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InterviewSolution
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If 80 g steam of temperature 97^@C is released on an ice slab of temperature 0^@C, how much ice will melt? How much energy will be transferred to the ice? When the steam will be transformed to water? |
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Answer» Solution :GIVEN : Latent HEAT of MELTING the ice = `L_"melt"`=80 cal/g Latent heat of vapourization of water =`L_"VAP"`=540 cal/g Solution:mass of steam =`m_"steam"`=80 g Temperature of steam=`97^@C` Temperature of ice = `T_"ice"=0^@C` Heatreleased during conversion of steam of temperature `97^@C` into water of temperature `97^@C= m_"steam" xx L_"vap"` =80 x 540 ...(i) Heat released during conversion of water of `97^@C` into water at `0^@C = m_"steam" xx DeltaT xx c` =80 x (97-0) x 1 = 80 x 97 ...(ii) Total heat gained by the ice = 80 x 540 + 80 x 97 ... from equation (i) and (ii) = 80 (540 + 97 ) = 80 x 637 = 50960 cal Some mass of the ice `(m_"ice")` , will melt due to this heat gained by the ice, then, `m_"ice"xxL_"melt"`=50960 cal `m_"ice" xx 80`=50960 `m_"ice"`=637 g Thus, 637 g ice will melt and 50960 cal will be given to the ice . |
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