# how do i find how much energy needs to be removed?

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

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

• 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 $n$.
– Zhe
Jan 17 '17 at 23:02
• On closer inspect, it's not even clean what $H$ is in that final equation.
– Zhe
Jan 17 '17 at 23:07

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.