Question

Solid substance A has speci c heat capacity 600 J kg?1K?1. When 2
kg of A at temperature 300 K is brought into thermal contact with 3 kg of
solid substance B at 240 K, the equilibrium temperature is 270 K. Find the
speci c heat capacity of substance B (assume no phase changes).

Answer 1

Given that final temperature is 270 K

Now using energy conservation:

Heat released by Solid A = Heat gained by Solid B

Qb = Qb

Ma*Ca*dT1 = Mb*Cb*dT2

Ma = Mass of solid A = 2 kg

Mb = Mass of solid B = 3 kg

Ca = specific heat capacity of A = 600 J/kg-K

Cb = specific heat capacity of B = ? J/kg-K

dT1 = 300 - 270 = 30 K

dT2 = 270 - 240 = 30 K

Using these values:

2*600*30 = 3*Cb*30

Cb = 2*600/3

Cb = 400 J/kg-K

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