The temperature of an iron part weighing 0.2 kg is 365 degrees. How much heat does it transfer to surrounding bodies

| July 2, 2021 | education

The temperature of an iron part weighing 0.2 kg is 365 degrees. How much heat does it transfer to surrounding bodies by cooling to 90 degrees at a temperature of 15 degrees.

To find the heat released during the cooling of the taken iron part, we apply the formula: Q = Czh * m * (tн – tх).

Constants and variables: Szh – specific heat of iron (Szh = 452 J / (kg * K)); m is the mass of the part (m = 0.2 kg); tн – temperature before cooling (tн = 365 ºC); t1 – first end temperature (t1 = 90 ºC); t2 – second end temperature (t2 = 15 ºC).

Calculation: a) Q1 = Czh * m * (tn – t1) = 452 * 0.2 * (365 – 90) = 24.86 * 10 ^ 3 J = 24.86 kJ.

b) Q2 = Czh * m * (tn – t2) = 452 * 0.2 * (365 – 15) = 31.64 * 10 ^ 3 J = 31.64 kJ.

Answer: An iron part when cooled to temperatures of 90 ºC and 15 ºC will transfer 24.86 kJ and 31.64 kJ of heat to the surrounding bodies, respectively.



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