5Kg de cobre que se encuentra en fase liquida a su temperatura de fusión, se la deja enfriar hasta temperatura ambiente de 15°C. ¿Cuántos KCal cedió al ambiente?

Datos Ce = 0.093 kcal/kg ºC (Se busca la capacidad calorífica del cobre en tablas) m = 5 kg t0 = 25 ºC (Temperatura ambiente estándar) t1 = 15 º Fórmula Q = m Ce (t1 - t0) Sustituyendo Q = (5 kg) (0.093 kcal / kg ºC) (15 - 25) ºC Q = (5 kg) (0.093 kcal / kg ºC) (- 10) ºC Q = - 4.65 kcal

If you go to this website: https://beta.openai.com/playground And you ask your question there, you will receive the following reply from an AI application that is currently in beta and not really ready for official publication: Assuming that the copper is in the form of molten copper, the heat released from the copper as it cools from its melting point (1084°C) to room temperature (15°C) can be calculated using the equation Q = m*Cp*ΔT, where m is the mass of copper (5 kg), Cp is the specific heat capacity of copper (0.385 J/g/K), and ΔT is the change in temperature (1084 - 15 = 1069 K). Q = (5 kg) x (0.385 J/g/K) x (1069 K) = 2037.7 KJ 1 Kcal = 4.184 KJ, so the heat released from the copper is 2037.7 KJ / 4.184 KJ/Kcal = 488.2 KCal.

Gracias a los dos ahora se como puedo resolverlo, aunque a los dos les dio resultado diferente lo entiendo, se como resolverlo, saludos.