InterviewSolution
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InterviewSolution
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If 80 g steam of temperature 97^@ Cis 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 stam will be transformed to water ? (Given : Latent heat of melting the ice ) L_("melt")=80 cal//g Latent heat of vaporiasation of water =L_("vap")=540 cal // g ) |
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Answer» Change is temperature `(DeltaT)` `=97-0=97^@C` We KNOW that: Latent heat of melting of ice `=L_("melt")=80 cal // g ` Latent heat of vaporisation of water `=L_("vap")=540 cal//g ` Specific heat of water `c_w = 1cal//g^@C` To find : i.Energy transferred (Q). iiMass of ice that melts (mi) FORMULAE: i . Heat released during conversion of steam into water at `97^@C(Q_1)= m_s xx L_(vap)` ii. Heat released during decrease of temperature of water from `97 ""^@C" "" to " " 0 "" "^@C " " (Q_2)` `=m_sxx c_s xx Delta T ` iii.Heat gained by ice (Q)`= m_i xx L_("melt")` Calculation From formula (i), `Q_1=80 xx 540 cal` From formula (ii) `Q_2= 80 xx 1 xx (97-0)` `= 80 xx 97 cal` According to princple of heat exchange , Total heat gained by ice `Q=Q_1+Q_2` `= 80 xx 540 +80xx 97` `=80 xx (540+97)` `= 80 xx 637` = 50960 cal This energy would cause `m_i` mass of ice to melt . From formula (iii), `therefore "" m_i xx L_("melt") = 50960` `therefore ""m_i =(50960)/(80) ` `=(80 xx 637)/(80)` `=637 g ` |
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