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:<br /><br />1. **Formula:** The formula for calculating heat transfer is: Q = mcΔT<br /><br /> Where:<br /> * Q = heat energy (Joules)<br /> * m = mass (grams)<br /> * c = specific heat capacity (J/g°C)<br /> * ΔT = change in temperature (°C)<br /><br />2. **Convert mass:** The mass is given in kilograms, but the specific heat capacity is in J/g°C. Convert the mass to grams:<br /><br /> 0.600 kg * 1000 g/kg = 600 g<br /><br />3. **Calculate the change in temperature (ΔT):**<br /><br /> ΔT = Final Temperature - Initial Temperature<br /> ΔT = 50.0°C - 100°C = -50.0°C (The negative sign indicates a decrease in temperature)<br /><br />4. **Plug the values into the formula:**<br /><br /> Q = (600 g) * (0.450 J/g°C) * (-50.0°C)<br /> Q = -13500 J<br /><br />5. **Interpret the result:** The negative sign indicates that heat is being *removed* from the iron. Therefore, 13,500 Joules of heat must be removed.<br /><br />**Answer:** 13,500 Joules of heat must be removed.<br />