Question

Do 3kg wody o temp. 10 stopni C wlano 3kg wody o temp.80 stopni C.Oblicz temp.końcową mieszaniny.

Answer 1

dane:

m_1 = 3kg

t_1 = 10C

m_2 = 3kg

t_2 = 80C

t_3 = ?

 

Q_1 = 4200\frac{J}{kg*C} * 3kg * ( t_3 - 10)C

Q_1 =  12600 (t_3 - 10) J

 

Q_2 = 4200\frac{J}{kg*C} * 3kg * (80 - t_3)C

Q_2 = 12600 (80 - t_3) J

 

Q_1 = Q_2

12600 (t_3 - 10) J = 12600 (80 - t_3) J

126t_3 - 1260 =  10080 - 126t_3

126t_3 + 126t_3 = 10080 + 1260

252t_3 =  11340/:252

t_3 =  45C^{o}