m1•c1•(100-24.1)= m2•c2•(24.1-16) + m3•v3•(24.1-16),
c1 = (m2•c2+ m3•v3)•8.1/m1•75.9 =
= 811 J/kg•degr =0.194 cal/g•degr
A piece of metal of mass 112g is heated 100 degree C.,and dropped into a copper calorimeter of mass 40g containing 200g of water at 16 derreC. Neglecting the heat loss, the specific het of the meatal is nearly,if the equilibrium temperature reached is 24.1degree C,
OPTIONS :
1.0.194 cal?gm degree C.
2.0.294 cal?gm degree C.
2 answers
0.194cal/gm°C