Question

1. the specific heat of iron is 0.473 j/g°c. how much energy is required to heat a 40.0 g sample of iron from 35.0°c to 75.0°c? (3pts)

heat = mass × specific heat × δt
where δt = t_final - t_initial
□ 757 j
□ 2080 j
□ 6690 j
□ 1320 j

Explanation:

Step1: Calculate temperature change

$\Delta t = t_{final} - t_{initial} = 75.0^\circ\text{C} - 35.0^\circ\text{C} = 40.0^\circ\text{C}$

Step2: Plug values into heat formula

$\text{heat} = \text{mass} \times \text{specific heat} \times \Delta t = 40.0\ \text{g} \times 0.473\ \text{J/g}^\circ\text{C} \times 40.0^\circ\text{C}$

Step3: Compute the final value

$\text{heat} = 40.0 \times 0.473 \times 40.0 = 756.8\ \text{J} \approx 757\ \text{J}$

Answer:

757 J