Question

How much total heat transfer is necessary to lower the temperature of 0.215 kg of steam from 129.5 °C to -31.5 °C, including the energy for phase changes? How much time is required for each stage of this process, assuming a constant 895.0 W rate of heat exchange? Give the times in the order that the stages occur.

Answer 1

Stage1) From 129.5 C to 100 C:

Time needed = 29.5 x 0.215 x 1996/895 = 14.1448 s

Stage2) For phase Change from steam to water:

Time needed = 0.215 x 2260000/895 = 542.9050 s

Stage3) From 100 C to 0 C:

Time needed = 100 x 0.215 x 4187/895 = 100.5816 s

Stage4) For phase Change from water to ice:

Time needed = 0.215 x 334000/895 = 80.2346 s

Stage5) From 0 C to -31.5 C:

Time needed = 31.5 x 0.215 x 2108/895 = 15.9513 s

Total heat exchange = 12659.63 + 485900 + 90020.5 + 71810 + 14276.43 = 674666.56 J

Total time = 753.8173 s