Question

27.0g of an unknown metal at 100 degrees Celcius was dropped into a beaker containing 313g of water initially at 22.3 degrees Celcius. The final temperature of the mixture was 25.1 degrees. What is the specific heat capacity of the unknown metal?

Answer 1

`1.81 J/g*^(o)C`

Step-by-step explanation:

`q = c_(p)*m*(T_(f) -T_(i))`

`q _(m)= c_(p)_(m)*27.0g*(25.1^(o)-100^(o))`

`\\\\q_(water) = c_(p)_{water*313 g*(25.1^(o) -22.3^(o))`

`q_(m) = - q_(water)\\\\c_(p)_(m)*27.0g*(25.1^(o)-100^(o)) = - 4.18 J/g*^(o)C*313 g*(25.1^(o) -22.3^(o))`

`c_(p)_(m)*2022.3 = 3663.352\\\\c_(pm)=1.81 J/g*^(o)C`