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A 0.0414 kg ingot of metal is heated to 243◦C
and then is dropped into a beaker containing 0.411 kg of water initially at 18◦C.
If the final equilibrium state of the mixed system is 20.4◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg ·◦ C.
Answer in units of J/kg ·◦ C.

Respuesta :

MathPhys

Answer:

448 J/kg/°C

Explanation:

m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0

(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0

(9.22 kg°C) C − 4129 J = 0

C = 448 J/kg/°C

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