Respostas

2014-03-29T16:02:22-03:00
Resposta

ΔT = 35ºC
Q = 3600cal

C =  \frac{Q}{T}

C = \frac{3600}{35}

C = 102,8cal/ºC

Calor especifico

c =  \frac{C}{m}

c = \frac{102,8}{500}

c = 0,2cal/g.C