Question

preciso da resposta do desafio 1) pra hj valendo 20 pontos!!!!

Answer 1

$Q_{ferro} + Q_{Agua} = 0 \\ \\ m.c.\Delta T + m.c.\Delta T = 0 \\ 300.c.(43 - 443) + 400.1.(43 - 10) = 0 \\ 300c.-400 + 400.33 = 0 \\ -120000c + 13200 = 0 \\ 120000c = 13200 \\ \\ c = \frac{13200}{120000} \\ \\ c = 0,11cal/g.C$