Commercial cold packs consist of solid ammonium nitrate and water.NH₄NO₃ absorbs 25.69 kJ of heat per mole dissolved in water.In a coffee-cup calorimeter,5.60 g NH₄NO₃ is dissolved in 100.0 g of water at 22.0°C.What is the final temperature of the solution? Assume that the solution has a specific heat capacity of 4.18 J/g⋅°C.

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Quizplus · Accepted Answer

First, convert the mass of ammonium nitrate (NH4NO3) to moles: $5.60 \, \text{g} \times \frac{1 \, \text{mol}}{80.04 \, \text{g/mol}} = 0.07 \, \text{mol}$. The heat absorbed when this amount dissolves is $0.07 \, \text{mol} \times 25.69 \, \text{kJ/mol} = 1.7983 \, \text{kJ} = 1798.3 \, \text{J}$. Since the heat is absorbed by the solution, it's lost by the water, so we use $q = mc\Delta T$ to find the temperature change, where $m = 100.0 \, \text{g} + 5.60 \, \text{g}$ (since the mass of the solution is the mass of the water plus the mass of the ammonium nitrate), $c = 4.18 \, \text{J/g°C}$, and $q = -1798.3 \, \text{J}$ (negative because the water loses heat). Solving for $\Delta T$, we find $\Delta T = \frac{q}{mc} = \frac{-1798.3 \, \text{J}}{(105.6 \, \text{g})(4.18 \, \text{J/g°C})} = -4.1°C$. The final temperature is $22.0°C - 4.1°C = 17.9°C$.

Commercial Cold Packs Consist of Solid Ammonium Nitrate and Water