Question

1. how much heat is needed to vaporize 25.0 g of water that is at -17.5°c to reach a final temperature of 115°c? (5 steps) 7.70 × 10^4 j

Explanation:

To solve the problem of finding the heat needed to vaporize 25.0 g of water from -17.5°C to 115°C, we follow these steps:

Step 1: Heat to raise ice to 0°C

The specific heat capacity of ice, \( c_{\text{ice}} = 2.09 \, \text{J/g°C} \). The temperature change \( \Delta T_1 = 0 - (-17.5) = 17.5 \, \text{°C} \).
The heat required, \( q_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1 \)
\( q_1 = 25.0 \, \text{g} \times 2.09 \, \text{J/g°C} \times 17.5 \, \text{°C} \)
\( q_1 = 25.0 \times 2.09 \times 17.5 \)
\( q_1 = 913.125 \, \text{J} \)

Step 2: Heat to melt ice at 0°C

The enthalpy of fusion of water, \( \Delta H_{\text{fusion}} = 334 \, \text{J/g} \).
The heat required, \( q_2 = m \cdot \Delta H_{\text{fusion}} \)
\( q_2 = 25.0 \, \text{g} \times 334 \, \text{J/g} \)
\( q_2 = 8350 \, \text{J} \)

Step 3: Heat to raise water to 100°C

The specific heat capacity of water, \( c_{\text{water}} = 4.184 \, \text{J/g°C} \). The temperature change \( \Delta T_2 = 100 - 0 = 100 \, \text{°C} \).
The heat required, \( q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 \)
\( q_3 = 25.0 \, \text{g} \times 4.184 \, \text{J/g°C} \times 100 \, \text{°C} \)
\( q_3 = 25.0 \times 4.184 \times 100 \)
\( q_3 = 10460 \, \text{J} \)

Step 4: Heat to vaporize water at 100°C

The enthalpy of vaporization of water, \( \Delta H_{\text{vaporization}} = 2260 \, \text{J/g} \).
The heat required, \( q_4 = m \cdot \Delta H_{\text{vaporization}} \)
\( q_4 = 25.0 \, \text{g} \times 2260 \, \text{J/g} \)
\( q_4 = 56500 \, \text{J} \)

Step 5: Heat to raise steam to 115°C

The specific heat capacity of steam, \( c_{\text{steam}} = 2.03 \, \text{J/g°C} \). The temperature change \( \Delta T_3 = 115 - 100 = 15 \, \text{°C} \).
The heat required, \( q_5 = m \cdot c_{\text{steam}} \cdot \Delta T_3 \)
\( q_5 = 25.0 \, \text{g} \times 2.03 \, \text{J/g°C} \times 15 \, \text{°C} \)
\( q_5 = 25.0 \times 2.03 \times 15 \)
\( q_5 = 761.25 \, \text{J} \)

Step 6: Total heat required

The total heat \( Q = q_1 + q_2 + q_3 + q_4 + q_5 \)
\( Q = 913.125 + 8350 + 10460 + 56500 + 761.25 \)
\( Q = 76984.375 \, \text{J} \)
Converting to scientific notation, \( Q \approx 7.70 \times 10^4 \, \text{J} \)

Answer:

The total heat needed is \( \boxed{7.70 \times 10^4 \, \text{J}} \)