An insulated cup contains 75g of water at 24 degrees. A 26g sample of metal at 82.25 degrees is added. The final temperature of the water and metal is 28.34 degrees. What is the specific heat of the metal?

Heat lost by the metal = Heat gained the mass of water

Now,

Heat = mass*change in temperature*specific heat capacity

Specific heat of water = 4.18 J/g.°C

Therefore,

75*4.18 * (28.34-24) = 26*C * (82.25-28.34) where C = Specific heat of the metal

C = (75*4.18*4.34) / (26*53.91) = 0.971 J/g.°C