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One of the moving parts of an engine contains 1.65 kgkg of aluminum and 0.300 kgkg of iron and is designed to operate at 211 ∘C∘C. How much heat is required to raise its temperature from 20.3 ∘C

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Final answer:

The total heat required to raise the temperature of 1.65 kg of aluminum and 0.300 kg of iron from 20.3°C to 211°C is 307.55 kJ, with 281.765 kJ for aluminum and 25.785 kJ for iron.

Step-by-step explanation:

To calculate the amount of heat required to raise the temperature of a substance, we use the formula Q = mcΔT, where 'Q' is the heat energy in joules, 'm' is the mass in kilograms, 'c' is the specific heat capacity in J/kg°C, and 'ΔT' is the change in temperature in degrees Celsius. For the engine part made of aluminum and iron, we need to calculate the heat required separately for each material and then add them together.

Heat Required for Aluminum

The specific heat capacity of aluminum is given as 900 J/kg°C. Thus, for 1.65 kg of aluminum:

QAl = (1.65 kg)(900 J/kg°C)(211 °C - 20.3 °C)

QAl = (1.65 kg)(900 J/kg°C)(190.7 °C) = 28,1765 kJ

Heat Required for Iron

Assuming the specific heat capacity of iron is approximately 450 J/kg°C (a value commonly used in similar calculations), for 0.300 kg of iron:

QFe = (0.300 kg)(450 J/kg°C)(211 °C - 20.3 °C)

QFe = (0.300 kg)(450 J/kg°C)(190.7 °C) = 25,785.15 J or 25.785 kJ

Total Heat Required

To find the total heat required, we add the heat required for both materials:

Total Q = QAl + QFe

Total Q = 281.765 kJ + 25.785 kJ = 307.55 kJ

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