How to Calculate Specific Heat of the metal?

How to calculate temperature change and specific heat of metal?

• the metal is copper the mass of metal is 57 g. start temp of water was 21 celsius, start temp of metal in water was 103 degrees celsius . final temp of water with metal in styrofoam cup is 26.1 degree celsius. total of three metals were placed in 100 degree boiling water. Later, they were taken out and placed in different styrofoam cups. hoe to find specefic heat of unknown metal by just the given data ?

• Answer:

  The solution to this is called the "method of mixtures". Basically, the idea is that the heat lost by the heated metal is equal to the heat gained by the cool water. Since the hot metal is dropped into the cool water, both the metal and the water end up at the same final temperature. One thing that simplifies the problem is that neither material changes phase during the exchange. There will be two equations that will be set equal to each other and then solved for the specific heat of the metal. a. Q(Cu) = m x (Ti - Tf) x Cp(Cu) b.) Q(H2O) = m (Tf-Ti) x Cp(H20) [m x (Ti-Tf) x Cp](Cu) = [m x (Tf-Ti) x Cp](H20) [57-g x (100 oC - 26.1 oC) x Cp](Cu) = [m x (26.1 oC - 21 oC) x 4.184-J/g* oC](water) [57-g x 73.9 oC x Cp](Cu) = [m x 5.1 oC x 4.184-j/g*oC](water) 4212.3Cp(Cu) = 21.34m(H20) because you didn't provide the mass of the cool water in the calorimeter cup, we can't go any farther with the calculation - we have 2 unknowns! Substitute in the mass of the water for 'm', multiply it by the 21.34 value then divide by the 4212.3 on the left and you will have the value for copper's specific heat. One other note - instead of the 103 o temp for the metal, I replaced it with the temperature of the boiling water. Since the metal was in the water, it cannot have a temperature higher than that of the water!