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How much energy must be removed to convert $8.49\ \mathrm{g}$ of water at $24.5\ ^{\circ}\mathrm{C}$ to an ice cube at $-12.0\ ^{\circ}\mathrm{C}$? Give your answer as a positive value in Joules since only positive energy can be removed. $$c_{\mathrm{water}} = 4.184\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$ $$c_{\mathrm{ice}} = 2.03\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$ $$\Delta H_{\mathrm{fusion, water}} = 6.01\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$

O am just learning this material, and I'm struggling on what formula to use. Is it

$$q= n(H_{\mathrm{final}} - H_{\mathrm{initial}})$$

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    $\begingroup$ Several points: 1. Generally speaking, figuring out the formula is not the way you want to go because you want to understand the underlying concept. Then, figuring out which formula is free and you can use it correctly. 2. Please define the non-obvious variables in your equation, namely $n$. $\endgroup$
    – Zhe
    Jan 17, 2017 at 23:02
  • $\begingroup$ On closer inspect, it's not even clean what $H$ is in that final equation. $\endgroup$
    – Zhe
    Jan 17, 2017 at 23:07

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Firstly let us understand that heat removed would be directly proportional to the decrease in temperature and also the mass. $$\delta Q\,\,\alpha \,\, m\delta T$$ Here the specific heat capacity appears as the constant of proportionality.
Therefore we have, $$\delta Q\,\,=\,\,mc\delta T$$ Now all you have to do is to use this formula to convert 8.49g of water at 24.5°C to water at 0°C.
But after this you need to use the concept of latent heat i.e. heat you need to remove to convert water at 0°C to ice at 0°C.As this value is given as 6.01 kJ/mol you will have to remove that much amount of heat for every mole of water(so you need to convert grams into moles).
Now what you have is 8.49 grams of ice at 0°C. Now just use the formula again to convert ice at 0°C to ice at -12°C (using the value of c of ice this time).
At the end you need to add all those heat energies and you are done !!
The sum of heat energies is the amount of energy you need remove from 8.49g water at 24.5°C to convert it to ice at -12°C.
Hope this helps.

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