exosloey
contestada

A 90-g aluminum calorimeter contains 390 g of water at a temperature of 20C. Then a 160 g piece of metal, initially at 305C is added to the calorimeter. The metal, water, and calorimeter reach an equilibrium temperature of 32 C. What is the specific heat capacity of the metal? The specific heat capacity of aluminum is 910 J/kg ∙ K, and of water is 4190J/kg*K.



430 J/kg · K


470 J/kg · K


350 J/kg · K


310 J/kg · K

Respuesta :

Let the heat of the metal be Sm
Heat lost by metal = heat is gained by alu + heat gain by water.
Sw × 0.160 × (305-32) = 0.090 × (32 -20) ×
910 + 0.390 × ( 32-20) × 4186
sw = 470 J/kg ×k
the specific heat for metal is 470J/kg×k 

Answer:

470 J/Kg.K

Explanation:

Let c be the specific heat capacity of metal.

According to the principle of caloriemetry

Heat lost by the hot body = Heat gained by the cold body

Heat lost by the aluminium = Heat gained by metal + calorimeter

mass of Aluminium x specific heat of Aluminium x rise in temperature  + mass of water x specific heat of water x rise in temperature = mass of metal x specific heat of metal x fall in temperature

0.09 x 910 x (32 - 20)  + 0.390 x 4190 x (32 - 20) = 0.16 x c x (305 - 32)

982.8 + 19609.2 = 43.68 c

c = 471.4 J/Kg.K

c = 470 J/Kg.K (nearly)

Otras preguntas