Suppose you want to cool 0.25 kg of cola (mostly water ),at 25"^(@)C by adding ice initially at -20"^(@)C How much ice should you add so that the final temperature will be 0"^(@)C with all the ice melt? Neglectthe heat capacity of the container of the container Specific heat of ice isheat of cola [4160 j/kgK]

Suppose you want to cool 0.25 kg of cola (mostly water ),at 25"^(@)C b

1.	Suppose you want to cool 0.25 kg of cola (mostly water ),at 25"^(@)C by adding ice initially at -20"^(@)C How much ice should you add so that the final temperature will be 0"^(@)C with all the ice melt? Neglectthe heat capacity of the container of the container Specific heat of ice isheat of cola [4160 j/kgK]
Answer» Solution :HEAT lost by cola is ``(Q_cola=m_cola s_colaDeltaT_cola)`` `=(0.25 kg)(4160 J//kg K)(25"^(@)C-0"^(@)C)`Let the masses of the required ice `m_ice` then the heat needed to WARM it from -`20"^(@)C` to `0"^(@)C` is `Q_1=m_ice S_ice DeltaT_ice` `=m_ice (2000J kg^(-1) k^-1)(0"^(@)C-(-20"^(@)C)` `m_ice (4.0xx10^(4)J kg^(-1))` The heat needed to melt this mass of ice is `Q_2=m_ice L_1=m_ice (3.34xx10^5J Kg^(-1))` Total energy gained by ice is `Q_1+Q_2=m_ice(3.74xx10^5J Kg^(-1)` Using heat lost =Heat gained we GET 26000 J =`m_ice(3.74xx10^5 jKg^-1)` or `m_ice =0.07 kg =70 g`

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