Question

An insulated cup contains 66.0g of water at 22.00oC. A 28.50g sample of metal at 95.25oC is added. The final temperature of the water and metal is 27.84oC. What is the specific heat of the metal

Answer 1

Answer:

Specific heat of metal of the metal is 0.8394J/g°C

Explanation:

The heat the water gain is the same losing for the metal. The equation is:

m(Metal)*ΔT(Metal)*S(Metal) = m(Water)*ΔT(Water)*S(Water)

Where m is mass: 66.0g water and 28.5g Metal

ΔT is change in temperature: (95.25°C-27.84°C) = 67.41°C for the metal and (27.84°C - 22.00°C) = 5.84°C for the water

And S is specific heat of water (4.184J/g°C) and the metal

Replacing:

28.5g*67.41°C*S(Metal) = 66.0g*5.84°C*4.184J/g°C

S(Metal) = 0.8394J/g°C

Specific heat of metal of the metal is 0.8394J/g°C