Question

An unknown piece of metal weighing 100. g is heated to 90.0 degrees Celcius. It is dropped into 250 g of water at 20.0 degrees Celcius. When equilibrium is reached, the temperature of both the water and piece of metal is 29.0 degrees Celcius. Determine the specific heat of the metal using the fact that the energy lost by the metal must equal the energy absorbed by the water. Assume that the heat capacity of the container, a Styrofoam cup, is negligible.

My answer: 1.543 J/g C

Answer 1

Use ΣmcΔT=0.
For metal,
mcΔT=100*c*(29-90)=6100c
For water,
mcΔT=250*4.184*(29-20)=9414
Applying ΣΔT=0
=>
100*c*(29-90)=250*4.184*(29-20)
Solving for c gives 1.543 J-g^-1-°C^-1 as you had it.

Answer 2

Correction: the answer is more appropriately given as 1.54 J/(g-°C) as all the original information is given to 3 significant figures.