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A 50.0-g sample of liquid water at 25.0°C is mixed with 35.0 g of water at 83.0°C. The final temperature of the water is __________°C.

(A) 54.0
(B) 282
(C) 48.9
(D) 166
(E) 27.4

Respuesta :

joseaaronlara

Answer:

The answer to your question is: T2 = 48.9 °C

Explanation:

Data

mass 1 = 50 g

Temperature 1 = 25°C

mass 2 = 35 g

Temperature 1 mass 2 = 83°C

Formula

                      m1C1(T2 - T1) = m2C2 (T2- T1)

C1 and C2 is the same because the substance is water

T2 is the same because water reach equilibrium.

Substitution

                     50(T2 - 25) = -35(T2 - 83)

                     50T2 - 1250 = -35T2 + 2905

                     50T2 + 35T2 = +2905 + 1250

                     85T2 = 4155

                     T2 = 4155 / 85

                     T2 =  48.9 °C                

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