Question

Total heat exchange = Qiron + Qwater = 0 Qiron = (462 J/K) + (T -...

Total heat exchange = Qiron + Qwater = 0

Qiron = (462 J/K) + (T - 352*C) and Qwater = (167.6 J/K) + (T - 20*C)

[(462 J/K) + (T - 352*C)] + [(167.6 J/K) + (T - 20*C)] =0

629.6 J/K + (T - 263.621) = 0

At the end they combine the two specific heats. I see that they simply added the 462 and 167.6 to get the 629.6, but I don't understand is how they got [T - 263.621] from [T - 352*C] + [T - 20*C]

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Answer #1

Solution heat lest by İron = heat gain by wa +er 482(352-T)ニ167, 6 (7-20) 2,757( 352-T) .-T-20 →

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