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Una muestra de agua de 250g de agua se caliento desde 10°C hasta 60°C calcula la cantidad de calor absorbido por el agua el calor especifico del agua es de 4186J/Kj.K

Sagot :

Lanuel

Answer:

Calor absorbido, Q = 52325 Joules

Explanation:

Dados los siguientes datos;

Masa = 250g to kg = 250/1000 = 0.25kg

Temperatura inicial, T1 = 10°C to Kelvin = 273 + 10 = 283K

Temperatura final, T2 = 60°C to Kelvin = 273 + 60 = 333K

Capacidad calorífica específica = 4186 j/kg.K.

Para encontrar la cantidad de calor necesaria;

La capacidad calorífica viene dada por la fórmula;

[tex] Q = mct [/tex]

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 333 - 283

dt = 50 K

Sustituyendo en la fórmula, tenemos;

[tex] Q = 0.25*4186*50[/tex]

[tex] Q = 52325 [/tex]

Calor absorbido, Q = 52325 Joules

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