A 0.80 x10^2 g mass of tungsten at 100.0°C is placed in 2.20 x 10^2 g of water at 20.0°C. ?

Question

A 0.80 x10^2 g mass of tungsten at 100.0°C is placed in 2.20 x 10^2 g of water at 20.0°C. The mixture reaches equilibrium at 21.6°C. Calculate the specific heat of tungsten.

_____J/kg·K

Answer 1

Heat gain of water = heat loss of tungsten

mass of water x sp heat water x temp gain = mass tungsten x sp ht tungsten x temp loss of tungsten

assume sp heat of water is 4200 J/kgK

so :-

2.2 x 4200 x 1.6 = 0.8 x c x 78.4

c = 2.2 x 4200 x 1.6/0.8 x 78.4 = 235.71J/KgK