[tex]q = m \times ce \times (tf - ti) \\ 2288.7 = 470.3 \times ce \times (57.3 - 24.3) \\ \frac{2288.7}{470.3} = ce \times 33 \\ 4.86 = ce \times 33 \\ ce = \frac{4.86}{30} \\ ce = 0.15[/tex]
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[tex]q = m \times ce \times (tf - ti) \\ 2288.7 = 470.3 \times ce \times (57.3 - 24.3) \\ \frac{2288.7}{470.3} = ce \times 33 \\ 4.86 = ce \times 33 \\ ce = \frac{4.86}{30} \\ ce = 0.15[/tex]
RESPUESTA :
Calor específico : 0.15
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