Final answer:
To warm 125g of iron from 23.5°C to 78.0°C, assuming a specific heat capacity of 0.449 J/g°C, you would require 3.063 kJ of energy.
Step-by-step explanation:
To calculate the number of kilojoules needed to warm 125g of iron from 23.5°C to 78.0°C, you would use the formula:
q = mcΔT, where:
q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (J/g°C)
ΔT = change in temperature (final temperature - initial temperature, in °C)
However, the specific heat capacity of iron is not provided in the question, and it is essential for the calculation. Typically, the specific heat of iron is around 0.449 J/g°C.
Assuming the specific heat capacity of iron is 0.449 J/g°C, the calculation would be:
q = (125g)(0.449 J/g°C)(78.0°C - 23.5°C)
q = 125g * 0.449 J/g°C * 54.5°C
q = 3,062.8125 J
Since we need to find the energy in kilojoules, we convert joules to kilojoules by dividing by 1000:
q = 3,062.8125 J / 1000 = 3.063 kJ (rounded to three decimal places)
Because the temperature increased, the iron absorbed heat, indicating that the heat is positive.