Question

14.17 - Phase Change and Latent Heat Part A How much heat transfer is necessary to raise the temperature of a 0.180-kg piece of ice from -20.0°C to 125°C, including the energy needed for phase changes? 558 kJ You are correct. Previous Tries Your receipt no. is 163-4010 ab Part B How much time is required for each stage, assuming a constant 0.810 kJ/s rate of heat transfer? (Make a graph of temperature versus time for this process to visualize the time scale.) Ice -20.0°C to ice 0°C 9.29 s Ice 0°C to water 0°C 3.01 s Water 0°C to 100°C 3.76 s Water 100°C to steam 100°C 20.3 s Steam 100°C to 125.0°C 0.341 s Sho 5.17 ܓܚܝ Submit Answer Incorrect. Tries 1/5 Previous Tries

Answer 1

Past A ice from = ML = heat transfer to raise temperature of -200 to 0°c mc.AT 0.180%2.04x(0-(-20) = 7.344 kJ heat transfer in phase change at occ 0:180 x 334 = 60.12 kJ hreat transfer to raise temperature 0°C to 100°c (water) mc.at 90.1809 4.18 2x(100-0) 75.276 kJ heat transfer in phase change at 1000 em L=0.180 x 2260 406.8 kJ heat transfer to raise temperature looc to 128°C (vapour cmCat = 0.180 x 1.996 x(125-100) -8.9685 KS Total heat transfer = 7.344 +60.12 +75.276+406.8 +8.9685 558J

Past B rate of heat transfer =0.810 kJ/S Ice -20°c to oc t 7.344 kJ 0.810 kJ/S 19.07 set Ice ooc to water oc t- 60.12 kJ 174.22 sec 0.810 kJ/S water ooc to 100°c t = 75.276 kJ 192.934 sec 0,810 KJ/S water 100°c to vapour 100°C 502:22 sec 01810 KJ/S Steam looc to 125.0°c + 8.9685 kg 0.810 KJ IS 407 sec t- 406.8ks