Question

1. how much heat is needed to freeze 75.0 g of water that is at 33.0°c? (2 steps) 35400 j

Explanation:

Step 1: Cool water to 0°C

The formula for heat loss (or gain) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is temperature change. For water, \( c = 4.184 \, \text{J/g°C} \), \( m = 75.0 \, \text{g} \), \( \Delta T = 33.0 - 0 = 33.0 \, \text{°C} \).
\( Q_1 = 75.0 \times 4.184 \times 33.0 \)
\( Q_1 = 75.0 \times 138.072 \)
\( Q_1 = 10355.4 \, \text{J} \)

Step 2: Freeze water at 0°C

The formula for heat of fusion (freezing) is \( Q = mL_f \), where \( L_f \) (latent heat of fusion for water) is \( 334 \, \text{J/g} \).
\( Q_2 = 75.0 \times 334 \)
\( Q_2 = 25050 \, \text{J} \)

Step 3: Total heat

Total heat \( Q_{\text{total}} = Q_1 + Q_2 \)
\( Q_{\text{total}} = 10355.4 + 25050 \)
\( Q_{\text{total}} = 35405.4 \, \text{J} \) (≈ 35400 J as given)

Answer:

The total heat needed is approximately \( \boldsymbol{35400 \, \text{J}} \) (or 35405 J).