Answer to Question #121576 in General Chemistry for Naveen

Question

Piece of metals weighs 59.047g was heated to 100.0 degrees C then was put into 100.0mL of water (initially at 23.7C) the metal water were allowed to come to an equilibrium temperature determined to be 27.8 degrees C. Assuming no heat lost to the environment calculate specific heat of metal show your work.

Answer 1

$q_{metal}=q_{water}$

$(mass)(\Delta t)(C_p)=(mass)(\Delta t)(C_p)$ $(59.047g)(100°C-72.2°C)(x)=(100g)(27.8°C-23.7°C)$ $(59.047g)(72.2°C)(x)=(100g)(4.1°C)$ .$4.184Jg^{-1}°C^{-1}$ .

$x=0.402Jg^{-1}°C^{-1}$