Application Calorimeter

[mehaksk]

I have a question that I found and I am a little bit confused: A 100.0 g sample of metal at 100.0°C is placed in a calorimeter containing 200.0 g of water at 25.0°C. The final temperature of the system stabilizes at 30.0°C. If the specific heat capacity of water is 4.18J/gC, , calculate the specific heat capacity of the metal. Assume no heat loss to the surroundings and that the calorimeter has negligible heat capacity.

[Rachel]

You're going to want to start by setting up mc(tf-ti) = mc(tf-ti) for the metal and the water. So you would have mass of the water (200) times specific heat of water, (4.18) times the final temperature minus the initial temperature, so ((30+273) - (25+273)) so one side of the equation looks like (200)(4.18)(5) = and the other side is the same but for the metal, and then you will leave the specific heat of the metal as x and then solve.