Question

When 565g of a metal with temperature of 90.00 C added to 200.0 g of water at 25.00 degrees. Assuming all the heat lost by the iron is transferred to the water ( specific heat capacity of water = 4/18 J/(g-C)), and the final temperature of the mixture is 34.40 degree Celsius what is the heat capacity of the metal?

Answer 1

The specific heat is 0.25J/g°C

Step-by-step explanation:

To solve this question we must know that the heat released for the metal is equal to the heat absorbed for the water. The equation is:

S(Metal)*m(Metal)*ΔT(metal) = S(water)*m(water)*ΔT(water)

Where S of metal is our incognite (Specific heat of metal)

m is the mass of the metal = 565g

ΔT is change in heat of the metal = 90.00°C - 34.40°C = 55.60°C

Specific heat of water = 4.18J/g°C

The mass of water = 200.0g

ΔT of water is 34.40°C - 25.00°C = 9.40°C

Replacing:

S(Metal)*565g*55.60°C = 4.18J/g°C*200.0g*9.40°C

S(metal) = 0.25J/g°C

The specific heat is 0.25J/g°C