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What-mass-of-ice-at-14-will-be-needed-to-cool-200-cm-3-of-an-orange-drink-essentially-water-from-25-C-to-10-C-specific-latent-heat-of-fusion-of-ice-3-36-10-5-jkg-1-specific-heat-capac




Question Number 6648 by Tawakalitu. last updated on 07/Jul/16
What mass of ice at  −14 will be needed to cool 200 cm^3  of  an orange drink (essentially water) from 25°C  to  10°C  (specific latent heat of fusion of ice = 3.36 × 10^5  jkg^(−1)   specific heat capacity of ice = 2100 jkg^(−1) k^(−1)   specific heat capacity of water = 4200 jkg^(−1) k^(−1) .
$${What}\:{mass}\:{of}\:{ice}\:{at}\:\:−\mathrm{14}\:{will}\:{be}\:{needed}\:{to}\:{cool}\:\mathrm{200}\:{cm}^{\mathrm{3}} \:{of} \\ $$$${an}\:{orange}\:{drink}\:\left({essentially}\:{water}\right)\:{from}\:\mathrm{25}°{C}\:\:{to}\:\:\mathrm{10}°{C} \\ $$$$\left({specific}\:{latent}\:{heat}\:{of}\:{fusion}\:{of}\:{ice}\:=\:\mathrm{3}.\mathrm{36}\:×\:\mathrm{10}^{\mathrm{5}} \:{jkg}^{−\mathrm{1}} \right. \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{ice}\:=\:\mathrm{2100}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{water}\:=\:\mathrm{4200}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} . \\ $$

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