Question

21. What is the enthalpy change (KJ) during the process in which 200.0 g of water at 60.0 °C is cooled to -25°C? The specific heats of ice, and liquid water, and 2.03 J/g-K, and 4.18 J/g-K, respectively. For H,O, AH = 6.01 kJ/mol A. 127 B. -127 C. 117 D. -117

Answer 1

200 gm of Water = 200/18 =11.11 moles of water

Change of temperature for the 1st process(60^oC---> 0^oC) is 60.

Q = m* T * Specific heat capacity of water

= 200 * 60 * 4.18 = 50160 J

change in enthalpy(H) = n * H_fus

= 11.11 * 6.01 = 66.77 KJ = 66770 J

Change in temperature for the 2nd process is ( 0^oC---> -25^oC) is 25.

Q = m* T * Specific heat capacity of ice

= 200 * 25 * 2.03 = 10150 J

Total energy released is (50160 + 66770 + 10150) = 127080 J = 127.08 KJ

Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings.

The enthalpy change is -127 KJ for the whole process.