$\mathbf{-}{\mathbf{q}}_{\mathbf{water}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{q}}_{\mathbf{ice}}$

Step 1

**${\mathbf{q}}_{\mathbf{water}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{mc}\mathbf{\u2206}\mathbf{T}\phantom{\rule{0ex}{0ex}}{\mathbf{q}}_{\mathbf{water}}\mathbf{}\mathbf{=}(360\mathrm{mL}\times \frac{0.998g}{1\mathrm{mL}})(4.18\frac{J}{g\xb7\xb0C})(0.0\xb0C-25.0\xb0C)$**

**q _{water }= -37544.76 J**

Use standard enthalpies of formation to calculate the standard change in enthalpy for the melting of ice. (The ΔH˚_{f} for H_{2}O(s) is –291.8 kJ/mol.) Use this value to calculate the mass of ice required to cool 360 mL of a beverage from room temperature (25.0˚C) to 0.0˚C. Assume that the specific heat capacity and density of the beverage are the same as those of water.

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