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suppose you want to cool 0.25 kg of cola...

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? Neglect the heat capacity of the container. specific heat of ice is` 2000 Jkg^(-1)K^(-1)`. [take specific heat of cola `4160 Jkg^(-1)K^(-1)`.]

Text Solution

Verified by Experts

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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