Question

A 0.050 0-kg ingot of metal is heated to 200.0 degree C and then dropped into a beaker containing 0.400 kg of water initially at 20.0 degree C. If the final equilibrium temperature of the mixed system is 22.4degree C, find the specific heat of the metal. How much heat is transferred to the water?

Answer 1

in the equlibrium condition we use energy conservation

heat released from the heated metal=heat is absorbed by water

$m_{metal}C_{metal}\Delta \Theta _{metal}=m_{water}C_{water}\Delta \Theta _{water}$

$0.05*C_{metal}(200-22.4)=0.4*4186*(22.4-20)$

$C_{metal}=452.54J/Kg^{\circ}C$

heat transfer to the water $H=mC\Delta \Theta$

   $H=0.4*4186*(22.4-20)$

H=4018.56 J