A44.0 g block of an unknown metal is heated in a hot water bath to have an initial temperature of 100.0°c. when the block is placed in a coffee-cup calorimeter containing 130.0 g of water initially at 25.0°c, the final temperature is 28.0°c. determine the specific heat of the unknown metal. the cs for water is 4.18 j/g°c.

Hey there!:

Q1 = -Q2

Q1 = m *Cp * ( Tf  - Ti )

Q1 = 44.0 * Cp * (28.0 - 100.0 )

Q2 = mw * CPw * ( Tf-Tw )

Q2 =  130 *4.18 * (28.0 - 25.0 )

Then :

-44.0*Cp*(  28.0 - 100.0 )  = 130 *4.18 * ( 28.0 -25.0 )

solve for Cp :

Cp = 130 *4.18* (28.0 -25.0) / ( 44.0 *( 100.0 - 28.0 )

Cp =  130 *4.18 * 3 / 44.0 * 72

Cp = 1630.2 / 3168

Cp =  0.5145 J/g°C

Hope that helps!

increased stability in the completion of fluorine's outer shell.