Question

determine a temperatura final de 50g de agua a 0 grau celsius forem adicionados a 250g de agua a 90 grau celsius num recipiente de capacidade termica dispinivel

Answer 1

Q = m*c*/\t

Q1 + Q2 = 0

50*1*(tf - 0) + 250 * (tf - 90) = 0 

50tf + 250tf - 22500 = 0

300tf = 22500

tf = 22500/300

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Tf = 75°C
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Answer 2

m.c.(tf - ti) + m.c.(tf - ti) = 0

50.c.(tf - 0) + 250.c.(tf - 90) = 0 ( simplifica o coeficiente)

50.(tf - 0) + 250.(tf - 90)

50 tf  +250tf - 22500 = 0

50tf + 250tf = 22500

300tf = 22500

tf = 22500/300

tf = 75°C

75° graus celsius