Question

1. how much heat is needed to boil 785 g of water that is at 17.6°c?

(2 steps) 2.04 × 10⁶ j

Explanation:

Step1: Heat to raise temperature to 100°C

We use the formula \( Q_1 = mc\Delta T \), where \( m = 785\space g \), \( c = 4.184\space J/g^\circ C \), and \( \Delta T = 100 - 17.6 = 82.4^\circ C \).
\( Q_1 = 785\times4.184\times82.4 \)
\( Q_1 = 785\times344.7616 \)
\( Q_1 = 270637.856\space J \)

Step2: Heat to vaporize water

We use the formula \( Q_2 = mL_v \), where \( L_v = 2260\space J/g \) (latent heat of vaporization of water).
\( Q_2 = 785\times2260 \)
\( Q_2 = 1774100\space J \)

Step3: Total heat

Total heat \( Q = Q_1 + Q_2 \)
\( Q = 270637.856 + 1774100 \)
\( Q = 2044737.856\space J \approx 2.04\times10^6\space J \)

Answer:

The heat needed is approximately \( 2.04\times10^6\space J \)