Question

A copper cylinder with a mass of 125 g and temperature of 345°C is cooled by dropping it into a glass beaker containing 565 g of water initially at 20.0°C. The mass of the beaker is 50.0 g and the specific heat of the glass is 840 J/kg∙K. What is the final equilibrium temperature of the system, assuming the cooling takes place very quickly, so that no energy is lost to the air? The specific heat of copper is 385 J/kg∙K and that of water is 4190 J/kg∙K.

Answer 1

Let the equilibrium temperature be T oC.

Now

Heat lost by Copper Cylinder = Heat gained by water + heat gained by the glass beaker.

Now heat gain or loss is given by the formula mxCpxdeltaT

Where m is the mass; Cp is the specific heat; delta T is the temperature change.

(0.125 kg)x(385 J/kg-K)x(345 - T) = (0.565 kg)x(4190 J/kg-K)x(T - 20) + (0.050 kg)x(840 J/kg-K)x(T - 20).

16603.125 + 48.125T = 2367.35T - 47347 + 42T - 840

64790.125 = 2361.225T

T = 27.44 ^oC.