Answer to Question #115193 in Molecular Physics | Thermodynamics for Sathu

Question

A 1.5 kg block of wood and a 2.5 kg block of steel are placed in thermal contact.
If the wood was at an initial temperature of 50°C and the steel was at an initial temperature of 15°C then what is the final equilibrium temperature of the wood and steel? (to 2 s.f and in °C)
[cwood = 2400 J kg−1 K−1, csteel = 490 J kg−1 K−1]

Answer 1

The final equilibrium temperature is T.

The wood loses heat, the steel gains the heat. The amount of heat in both cases is the same according to equilibrium condition.

$Q(loss) = Cm(T_w - T) = 2400*1.5*(50-T) = 180000 - 3600T$

$Q(gain) = Cm(T-T_s) = 490*2.5*(T-15) = 1225T - 18375$

$Q(loss) = Q(gain)$

$180000-3600T = 1225T - 18375$

$4825T = 198375$

T = 41.11 ^oC -- the final equilibrium temperature of the wood and steel