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

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

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