Questions: When a hot piece of copper is added to 45.5 g of 22.0°C water, their final temperature is 35.0°C. What is the heat gained by the water?

Solution Steps

We are given the following information:

• Mass of water, \( m = 45.5 \, \text{g} \)
• Initial temperature of water, \( T_i = 22.0^\circ \text{C} \)
• Final temperature of water, \( T_f = 35.0^\circ \text{C} \)
• Specific heat capacity of water, \( c = 4.184 \, \text{J/g}^\circ \text{C} \)

The change in temperature (\(\Delta T\)) of the water is calculated as: \[ \Delta T = T_f - T_i = 35.0^\circ \text{C} - 22.0^\circ \text{C} = 13.0^\circ \text{C} \]

The heat gained by the water (\(q\)) can be calculated using the formula: \[ q = m \cdot c \cdot \Delta T \] Substituting the known values: \[ q = 45.5 \, \text{g} \times 4.184 \, \text{J/g}^\circ \text{C} \times 13.0^\circ \text{C} \]

\[ q = 45.5 \times 4.184 \times 13.0 = 2475.292 \, \text{J} \]

Final Answer