Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

Answers

Answer 1
Answer:

Final answer:

Using the formula q = mcΔT, and substituting the values for mass, specific heat capacity of iron, and temperature change, it is calculated that it takes approximately 3.058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.

Explanation:

To calculate the amount of heatneeded to warm 125 g of iron from 23.5 °C to 78.0 °C, we use the formula q = mcΔT, where 'm' is the mass in kilograms, 'c' is the specific heat capacity, and 'ΔT' is the temperature change. In this case, the mass 'm' is 0.125 kg (since 1 g = 10^-3 kg), the specific heat capacity 'c' of iron is 0.449 J/g°C (or 449 J/kg°C), and 'ΔT' is 78.0 °C - 23.5 °C = 54.5 °C.

Substituting these values into the formula, we get q = (0.125 kg) * (449 J/kg°C) * (54.5 °C), which gives a result of approximately 3.058 KJ.

Therefore, it would take approximately 3,058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.

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Answer 2
Answer:

Final answer:

To warm 125 g of iron from 23.5 °C to 78.0 °C, it requires approximately 3.93 kilojoules of energy.

Explanation:

To calculate the number of kilojoules required to warm 125 g of iron from 23.5 °C to 78.0 °C, we can use the formula:

q = m * c * ΔT

Where:

  • q is the energy absorbed or released
  • m is the mass of the substance
  • c is the specific heat capacity of the substance
  • ΔT is the change in temperature

Using the given values:

  • m = 125 g
  • c = 0.450 J/g°C (specific heat capacity of iron)
  • ΔT = 78.0 °C - 23.5 °C

Substituting the values into the formula:

q = 125 g * 0.450 J/g°C * (78.0 °C - 23.5 °C)

Simplifying the equation:

q = 125 * 0.450 * (78.0 - 23.5)

q ≈ 3933.75 J ≈ 3.93 kJ

Therefore, it requires approximately 3.93 kilojoules of energy to warm 125 grams of iron from 23.5 °C to 78.0 °C.

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Answers

Answer:

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Explanation:

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