Is the 650 grams the solvent-water or cold pack? It's the water temperature that drops, not the mass of cold pack. So, I'm assuming the mass value is that of water into which the cold pack was added. Also, when measuring changes of temperature, one is actually measuring the change occurring in the surroundings not the system. The results 'observed' are typically interpreted in terms of the system (in this case the cold pack).
So, using the 650g as the mass of water-solvent losing heat to the dissolution of the cold pack salts, Q = mcΔT = (650g)(4.184j/g∙⁰C)(-15 - 24)⁰C = - 106,064.4 Joules energy lost from solvent-water, or -1.10E+5 Joules (2-Sig.Figs.)… which means that the cold pack material on dissolution gained (endothermic) the 1.1 x 10⁵ Joules.
That is, the reaction is exothermic with respect to the water-solvent but endothermic with respect to the cold pack.
The enthalpy of solution is the amount of energy absorbed or released when a solute is dissolved in a solvent. Suppose a 650 g ‘cold pack’ drops in temperature from 24 Celsius to -15 Celsius when its constituents mix. Assuming the specific heat of this aqueous solution is close to that of pure water, 4.18 J/g-C, what is the enthalpy of solution for the cold pack’s components? Is this enthalpy a positive or negative number? Explain.
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