Questions: What is the heat change when 2.42 g of water vapor (steam) at 189.3°C is cooled to 106.1°C? The specific heat of steam is 2.02 J/g°C. Be sure your answer has the correct number of significant figures.

Solution Steps

We are given:

• Mass of water vapor, \( m = 2.42 \, \text{g} \)
• Initial temperature, \( T_i = 189.3^\circ \text{C} \)
• Final temperature, \( T_f = 106.1^\circ \text{C} \)
• Specific heat of steam, \( c = 2.02 \, \frac{\text{J}}{\text{g}^\circ \text{C}} \)

The change in temperature, \(\Delta T\), is calculated as: \[ \Delta T = T_f - T_i = 106.1^\circ \text{C} - 189.3^\circ \text{C} = -83.2^\circ \text{C} \]

The heat change, \( q \), can be calculated using the formula: \[ q = m \cdot c \cdot \Delta T \] Substituting the given values: \[ q = 2.42 \, \text{g} \times 2.02 \, \frac{\text{J}}{\text{g}^\circ \text{C}} \times (-83.2^\circ \text{C}) \]

\[ q = 2.42 \times 2.02 \times (-83.2) = -406.1 \, \text{J} \]

Final Answer