Question

20. A 0.145 kg mass of tungsten at 130.0 °C is placed in a 0.502 kg of water at 22.0 °C. The mixture reaches equilibrium at 28.6 °C. Calculate the specific heat of tungsten. (specific heat of water = 4180 J/kg C)

Givens:


Unknown:


Equation: macaΔTa = - mbcbΔTb


Substitute:


Solve:

Answer 1

MaCa∆Tb = MbC∆Tb
Where m is the mass of a compound
C is the specific heat of a compound
∆T is the change of temperature of the compound
(0.502 kg water)(4180 j/kg C)( 28.6 – 22 C) = ( 0.145 kg tungsten) ( C ) ( 130 – 28.6 C)
Solve for C
C = 941.9 J /kg C