When 35 g of a metal at 94 ◦c is added to 51 g of water at 22 ◦c, the temperature of the water rises to 27 ◦c. what is the specific heat capacity of the metal?

M = mass of metal 
M = mass of water 
c = specific heat 
c of water = 1 cal/g *C 

use the formula 

Q = mcΔT = McΔT 
(27g)*(c)*(107 C - 27.3 C) = (54.2g)*(1 cal/g *C)*(27.3 C - 22 C) 
2152 * c = 287.26 
c = specific heat of metal = .133 cal/g *C 

1 cal / 4.186 j/g *C = .133 cal / x 
= .557 J/g *C

27. =3,500,000 x 4,000 = 14,000,000,000