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A thermally insulated pot has 150 g ice ...

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`

Text Solution

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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`
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