Question #160796

What total heat is released when 0.500 lb of steam at 212°F changes to ice at -6°F?


1
Expert's answer
2021-02-03T02:47:21-0500

Answer

Total heat is released

Q=msΔTQ= ms\Delta T

=4.59×0.500×2.04×121.11=567.02J=4.59\times0.500\times2.04\times 121.11\\=567.02J


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