A thermally insulated pot has 150 g ice at temperature 0^@C. How much steam of 100^@C has to be mixed to it, so that water of temperature 50^@C will be obtained? Given:latent heat of melting of ice (L_f) = 80 cal/g , latent heat of vapourization of water (L_v)=540 cal/g , specific heat of water (C_w)=1 cal//g^@ C

A thermally insulated pot has 150 g ice at temperature 0^@C. How much

1.	A thermally insulated pot has 150 g ice at temperature 0^@C. How much steam of 100^@C has to be mixed to it, so that water of temperature 50^@C will be obtained? Given:latent heat of melting of ice (L_f) = 80 cal/g , latent heat of vapourization of water (L_v)=540 cal/g , specific heat of water (C_w)=1 cal//g^@ C
Answer» Solution :The heat given of the stream `Q_S` will be equal to , `Q_S=m_S L_V + m_S c_W DeltaT` …(i) The heat taken in by the ice `Q_i` will be equal to , `Q_i=m_iL_f + m_ic_WDeltaT` ….(ii) Heat given off the stream `Q_s` is equal to heat taken in by the ice `Q_i`. So, `Q_s=Q_i` `m_SL_V+m_SC_W DeltaT = m_iL_f+m_ic_WDeltaT` [from (i) and (ii)] `(m_S xx 540 ) + (m_S xx 1 xx (100-50) = (150 xx 80) + {150 xx 1 xx (50-0)` `therefore m_S(540+50) = 12000 + 7500` `therefore 590 m_S = 19500` `therefore m_S=19500/590` `m_S`=33 G From the above observation we conclude that , 33 g of steam at `100^@` C MUST be MIXED with 150 g ice at `0^@C`