7. How much heat must be removed from 0.600kg of iron to lower its temperature from 100^circ C to 50.0^circ C (c_(lron)=0.450J/g^circ C)

Here's how to calculate the heat removed from the iron:1. **Formula:** The formula for calculating heat transfer is: Q = mcΔT Where: * Q = heat energy (Joules) * m = mass (grams) * c = specific heat capacity (J/g°C) * ΔT = change in temperature (°C)2. **Convert mass:** The mass is given in kilograms, but the specific heat capacity is in J/g°C. Convert the mass to grams: 0.600 kg * 1000 g/kg = 600 g3. **Calculate the change in temperature (ΔT):** ΔT = Final Temperature - Initial Temperature ΔT = 50.0°C - 100°C = -50.0°C (The negative sign indicates a decrease in temperature)4. **Plug the values into the formula:** Q = (600 g) * (0.450 J/g°C) * (-50.0°C) Q = -13500 J5. **Interpret the result:** The negative sign indicates that heat is being *removed* from the iron. Therefore, 13,500 Joules of heat must be removed.**Answer:** 13,500 Joules of heat must be removed.