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?

Explanation

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`
Heat released 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 .