What is the final temperature when 50 mL of water at 80°C are added to 25 mL of water at 25°C?

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Answer 1 · 2017-06-01T22:40:59

$61.7^oC$

$sum$ energy changes = 0

$sumQ = Q_(80^oC) + Q_(25^oC)$ = $[mcDeltaT_(80^oC) + mcDeltaT_(25^oC)$ ] = 0

Density HOH = 1.00 g/ml;
Speific Heat(c) HOH = $1.00 "cal"/g^oC$

$[(50g)(1("cal"/g^oC))(T_("final") - 80^oC)]$ +

$[(25g)(1("cal"/g^oC))(T_("final") - 25^oC)]$ = $0$

=> $[(50)(1)(T_("final") - 80)] + [(25)(1)(T_("final") - 25)] = 0$

=> $(50T_("final") - 4000) + (25T_("final") - 625) = 0$

=> $75T_("final") = (4000 + 625) = 4625$

=> $T = (4625/75)^oC = 61.7^oC$