A 0.600-kg piece of metal X is heated to 100°C and placed in an aluminum can of mass 0.200-kg which contains 0.500 kg of water initially at 17.3°C.The final equilibrium temperature of the mixture is 20.2°C,what is the specific heat of metal X? The specific heats of water and aluminum are 4186 J/kg ∙ K (water)and 910 J/kg ∙ K (aluminum).
A)140 J/kg ∙ K
B)270 J/kg ∙ K
C)450 J/kg ∙ K
D)900 J/kg ∙ K

Question

Quizplus · Accepted Answer

Explanation:Heat lost by metal X = Heat gained by water + Heat gained by aluminum can$$mC_p(T_f-T_i)=mC_p(T_f-T_i)+mC_p(T_f-T_i)$$$$0.6C_p(20.2-100)=0.5(4186)(20.2-17.3)+0.2(910)(20.2-17.3)$$$$-47.4C_p=14655.3+539.4$$$$C_p=\frac{15194.7}{47.4}=320.58\approx 320 J/kg \cdot K$$

A 0600-Kg Piece of Metal X Is Heated to 100°C and and Placed