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What is the enthalpy change (KJ) during the process in which 200.0 g of water at...

What is the enthalpy change (KJ) during the process in which 200.0 g of water at 60.0 Celsius is cooled to -25 Celsius? The specific heats of ice, and liquid water, and 2.03 J/g-K, and 4.18 J/g-K, respectively. For H 2 O, delta H fus = 6.01 kJ/mol

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Answer #1

The enthalpy change during the freezing process is given by

enthalpy change = (msdDT)WATER + (n*DHfusion) + (m*s*DT)ice

                   = (200*4.18*(60-0)+ (200/18)*(-)6.01*10^3) + 200*2.03*(0-(-25))

                   = -6.4677 KJ

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