an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the temperature of 98 c. is the system reaches thermal equilibrium at 78C determine th despicfic heat of the metal

The specific heat capacity of the metal is 0.843J/g°C

Explanation:

Hello,

To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.

Heat gained by the metal = heat loss by water + calorimeter

Data,

Mass of metal (M1) = 512g

Mass of water (M2) = 325g

Initial temperature of the metal (T1) = 15°C

Initial temperature of water (T2) = 98°C

Final temperature of the mixture (T3) = 78°C

Specific heat capacity of metal (C1) = ?

Specific heat capacity of water (C2) = 4.184J/g°C

Heat loss = heat gain

M2C2(T2 - T3) = M1C1(T3 - T1)

325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)

1359.8 × 20 = 512C1 × 63

27196 = 32256C1

C1 = 27196 / 32256

C1 = 0.843J/g°C

The specific heat capacity of the metal is 0.843J/g°C

hloride, what is the mass of calcium in

explanation: