Question

A metal bar with a mass of 4 kg is placed in boiling water until its temperature stabilizes at 100 degrees Celsius. The bar is then immersed in 500mL of water with an initial temperature of 20 degrees Celsius. The mixture reaches a temperature of 35 degrees celsius. What is the specific heat capacity of the metal bar?

Answer 1

The change in enthalpy of the water should equal the change in enthalpy of metal bar.

Specific heat capacity of water (c) is 4.18J/g/mol.
Δh = mcΔT
m = 100g
ΔT = 35 - 20 = 15degrees
Δh = 100*4.18*15 = 6270J

This is also Δh of metal bar.
m = 4kg = 4000g
ΔT = 100 - 35 = 65
c = Δh/(mΔT) = 6270/(4000*65) = 0.024J/g/mol

Answer 2

fitst off specific heat of water is 1 cal/g degree C
density of water is 1g/ml so 500mL = 500g
4kg = 4000g
since metal is in the water, their temperatures will even out in the end
them changes must be calculated in a way to get positive values so:
temp change of metal is 100 - 35 = 65C
Temp change of water is 35 - 20 = 15C

now plug into this thermochem formula:
1 cal/gxC x 500g x 15C (change in temp of water) = ________ x 4000g x 15C (change in temp of metal)
so the specific heat of the metal bar is .02884 cal/g degree C = .03 cal/g degree C (answer taking sig figs into account)